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UNIVERSITY   OF   CALIFORNIA. 

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Accessions  No. *$£-&-<r-^--3      Shelf  No. 


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STRENGTH 


BEAMS  AND  COLUMNS; 


IN    WHICH 

THE    ULTIMATE    AND    THE    ELAS- 
TIC  LIMIT  STRENGTH    OF   BEAMS    AND    COL- 
UMNS   IS   COMPUTED   FROM   THE    ULTIMATE   AND  ELASTIC 
LIMIT  COMPRESSIVE   AND   TENSILE   STRENGTH   OF  THE  MATERIAL, 
BY   MEANS    OF  FORMULAS  DEDUCED    FROM    THE   CORRECT 
AND    NEW   THEORY    OF  THE   TRANSVERSE 
STRENGTH     OF     MATERIALS. 

BY 

EOBEET   H.   COUSIKS, 

Civil  Engineer, 

Formerly  Assistant  P)'ofesaor  of  Mathematics  at  the  Virginia  Military  Institute, 
Lexington.,  Va. 


E.   &   F.    N.    SPON", 


NEW   YORK : 
12   CORTLANDT    STREKT. 


LONDON : 


125   STRAND. 


1889. 

[All  rights  reserved.] 


COPYRIGHT,  1889, 

BY 
R.   H.    COUSINS. 


INTRODUCTION. 


FOK  more  than  two  centuries  the  mathematical  and  me- 
chanical laws  that  govern  the  transverse  strength  of  Beams  and 
Columns  have  received  the  attention  of  the  most  expert 
mathematicians  of  all  countries.  Galileo,  in  1038,  formulated 
and  published  the  first  theory  on  the  subject.  He  was  fol- 
lowed by  such  philosophers  as  Mariotte,  Leibnitz,  Bernoulli!, 
Coulomb,  and  others,  each  amending  and  extending  the  work 
of  his  predecessor,  until  the  year  1824,  when  Navier  succinctly 
stated  the  theory  that  is  recognized  to  be  correct  at  the  present 
day,  and  to  which  subsequent  writers  and  investigators  have 
added  but  little. 

This  theory  has  neither  received  the  endorsement  of  the 
experimenters  nor  of  some  of  the  theoretical  writers.  * "  Ex- 
cepting as  exhibiting  approximately  the  laws  of  the  phenom- 
ena, the  theory  of  the  strength  of  materials  has  many  prac- 
tical defects"  (Wiesbach).  "  It  has  long  been  known  that  under 
the  existing  theory  of  beams,  which  recognizes  only  two  ele- 
ments of  strength  —namely,  the  resistance  to  direct  compression 
and  extension — the  strength  of  a  bar  of  iron  subjected  to  a 
transverse  strain  cannot  be  reconciled  with  the  results  obtained 
from  experiments  on  direct  tension,  if  the  neutral  axis  is  in 
the  centre  of  the  bar"  (Barlow). 

During  the  present  century  much  time  and  means  have 
been  expended  in  attempts  to  solve,  experimentally,  the  prob- 
lems that  have  engaged  the  attention  of  the  mathematicians, 
and  as  the  result  of  their  labors  we  find  such  experimenters  as 
Hodgkinson,  Fairbairn,  and  others,  whose  names  are  house- 


IV  INTRODUCTION 

hold  words  in  the  literature  of  the  subject,  adopting  empirical 
rules  for  the  strength  of  .Beams  and  Columns  rather  than  the 
rational  formulas  deduced  by  the  scientists.  "  For  no  theory 
of  the  rupture  of  a  simple  beam  has  yet  been  proposed  which 
fully  satisfies  the  critical  experimenter'-  (De  Yolson  Wood). 

That  we  should  be  able  to  deduce  the  strength  of  Beams  and 
Columns  from  the  known  tensile  and  compressive  strength  of 
the  material  composing  them,  has  been  apparent  to  many 
writers  and  experimenters  on  the  subject,  but  to  the  present 
time  no  theory  has  been  advanced  that  embodies  the  mathe- 
matical and  mechanical  principles  necessary  to  its  accomplish- 
ment. The  theory  herein  advanced  and  the  formulas  resulting 
therefrom  deduce  the  strength  of  Beams  and  Columns  from 
the  direct  crushing  and  tensile  strength  of  the  material  com- 
posing them,  without  the  aid  of  that  coefficient  that  has  no 
place  in  nature,  the  Modulus  of  Rupture.  The  theory  and 
the  formulas  deduced  therefrom  are  in  strict  accord  with  cor- 
rect mechanical  and  mathematical  principles,  and  the  writer 
believes  that  they  will  fully  satisfy  the  results  obtained  by  the 
experimenter. 

The  great  practical  benefits  to  be  derived  from  the  correct 
theory  of  the  strength  of  Beams  and  Columns  will  be  evident, 
when  we  consider  the  countless  tons  of  metal  that  have 
been  made  into  railroad  rails,  rolled  beams,  and  the  other 
various  shapes,  and  that  the  manufacturers  were  without 
knowledge  of  the  work  to  be  performed  by  the  different  parts 
of  the  beam  or  column  in  sustaining  the  load  that  it  was  in- 
tended to  carry.  The  best  that  they  have  been  able  to  do  is 
to  compute  the  strength  by  the  aid  of  an  empirical  quantity 
deduced  from  experiments  on  "  similar  beams."  The  correct 
theory  will  enable  them  to  foretell  the  strength  of  any  untried 
shape,  and  the  reason  for  the  strength  of  those  that  have  been 
long  in  use,  which  is  the  "  true  object  of  theory." 

E.  II.  C. 

DALLAS,  TEXAS,  March  13,  1889. 


CONTENTS. 


CHAPTER  I. 

FORCES   DEFINED    AND   CLASSED. 

ART.  PAGE 

1.  Force  Defined, 1 

2.  Stress  or  Strain,          .........  1 

3.  The  Load, 2 

4.  Equilibrium  and  Resultant.        ....*...  2 

5.  Bending  Moment — Concentrated  Forces,       .....  3 

6.  General  Formulas  for  Bending  Moments,           ....  6 

7.  Uniformly  Varying  Forces— Rectangular  Areas,           ...  8 

8.  Resultant,            8 

9.  Moment  of  Uniformly  Varying  Forces,         .         .         .         •*      •  9 

10.  "        "            "              "              "  11 

11.  Uniformly  Varying  Forces — Circular  Segment  Areas,           ,         .  13 

12.  Moment  of  Uniformly  Varying  Forces — Circular  Segment  Areas,  13 

13.  ""              '           '    "              "               "              "  14 

14.  Resultant,              "                             "  *  16 

15.  Moment  of  Uniformly  Varying  Forces — Circular  Arcs,        .         .  17 

16.  '.'        "            "                             V                            "  18 

17.  Resultant,             "                                              "            "  1-9 


CHAPTER  II. 

RESISTANCE    OF    CROSS  SECTIONS   TO    RUPTURE. 

18.  Moment  of  Resistance, 21 

19.  Neutral  Line, 21 

20.  Bending  JL  and  Moment  of  Resistance,    .  22 

21.  Equilibrium, 23 

22.  Position  of  the  Neutral  Line 24 

23.  Neutral  Line  at  the  Transverse  Elastic  Limit,       ....  24 

24.  Movement  of  the  Neutral  Line,  with  the  Deflection,         .         .  26 

25.  Neutral  Lines  of  Rupture  in  a  Rectangular  Section,      ...  28 

26.  Relative  Value  of  the  Coinpressive  and  Tensile  Strains,  28 

27.  Summary  of  the  Theory, 31 


VI  CONTENTS. 

CHAPTER  Til. 

TRANSVERSE   STRENGTH. 

ART.  PAGE 

28.  Coefficients  of  Strength, 33 

29.  Elasticity  of  Materials,    .        .        .        ...        .        .        .33 

30.  Elastic  Limits,            34 

31.  Elastic  Limit  of  Beams, 34 

32.  Working  Load  and  Factors  of  Safety, 34 

33.  General  Formula — Transverse  Strength,        .         ..":•'..         .         .35 

34.  Relative  Transverse  Strength  of  a  Beam, 35 

35.  Moment  of  Resistance — Rectangular  Sections,       ....  37 

36.  The  Neutral  Line                                                 ....  38 

37.  Transverse  Strength                "                "               ....  38 

38.  Designing  a  Beam                     "                "          ....  39 

39.  To  Compute  the  Compressive  Strain, 39 

39.  "                     "    Tensile                              39 

40.  Moment  of  Resistance — Hodgkinson  Sections,       .         .         .         .40 

41.  Neutral  Line                               "                "          .         .         ...  42 

42.  Transverse  Strength                  "                               ....  44 

43.  To  Design  a  Hodgkinson  Section, 44 

44.  Moment  of  Resistance— Dbl.  T  and  Hollow  Rectangular  Sections,  47 

45.  Neutral  Line                       "'-•'••.'«                         •«  4$ 

46.  Transverse  Strength                                                "                   «<  4# 

47.  To  Design  a                         "         "          "           '  »                   <•  48 

48.  Moment  of  Resistance— In vtd  T,  Dbl.  Invtd.  T  and  U  Sections,  52 

49.  Neutral  Line                           "  53 

50.  Transverse  Strength  53 

51.  To  Design  an  53 

52.  Moment  of  Resistance— Circular  Sections, 55 

53.  Neutral  Line                          "              "  56 

54.  Transverse  Strength             "                           56 

55.  To  Compute  the  Compressive  Strain  of  Circular  Sections,         .  58 

56.  "                      "   Tensile               ««,«««                           .        .  58 

57.  Relative  Strength  of  Circular  and  Square  Sections,  .         .         .  59 

58.  Moment  of  Resistance— Hollow  Circular  Sections,        .         .         .60 

59.  Neutral  Line                                                       "  60 

60.  Transverse  Strength            "                             "  60 

CHAPTER  IV. 

CAST-IRON    BEAMS. 

61.  Compressive  Strength  of  Cast-Iron,    .  64 

62.  Tensile                 "         "          "  64 


CONTENTS.  Vll 

ART.  PAGE 

63.  Ratio  of  the  Compressive  to  the  Tensile  Strength,     ...  65 

64.  Transverse  Strength  of  Cast-Iron,          " 65 

65.  To  Compute  the  Compressive  Strength  of  Cast-iron,         .         .  67 

65.  "  "         "     Tensile  "  ...       67 

66.  Neutral  Line— Rectangular  Cast-Iron  Beams,   ....  71 

67.  Transverse  Strength     "  •'  ....       71 

68.  To  Design  a  Rectangular  Cast-iron  Beam,         ....  73 

69.  Hodgkinson  Beams,       .  .        .        .        .        .        .        .74 

70.  Mr.  Hodgkinson 's  Experiments,         ......  74 

71.  Neutral  Line— Hodgkinson  Cast-iron  Beams,        .         .         .         .75 

72.  Transverse  Strength    "  "  "  75 

73.  To  Design  a  Hodgkinson  Cast-Iron  Beam, 78 

74.  Neutral    Line— Double  T  and  Box  Cast-Iron  Beams,        .         .  78 

75.  Transverse  Strength    "         "  "  "         .         .         .78 

76.  To  Design  a  Double  T  and  Box  Cast-iron  Beam,      .        .    •     .  79 

77.  Neutral  Line— Cast-Iron  Circular  Beams, 80 

78.  Transverse  Strength  "  "  "  ....  81 

79.  Movement  of  the  Neutral  Line 82 

80.  Relative  Strength  of  Circular  and  Square  Cast-Iron  Beams,      .  83 

81.  Neutral  Line — Hollow  Circular  Cast  Iron  Beams,        .         .         .85 

82.  Transverse  Strength  '•  «  ...  86 

CHAPTER  V. 

WIIOUGHT-IRON    AND    STEEL   BEAMS. 

83.  Compressive  Strength  of  Wrought-Iron,         .         .  87 

84.  Tensile                  .«          «.                              87 

85.  Compressive        "          "  Steel 88 

86.  Tensile                             "      "             . 88 

87.  To  Compute  Compressive  Strength  of  Wrought-Iron  and  Steel,  88 

88.  "  "        Tensile  "  90 

89.  Transverse  Strength  of  Wrought-Iron  Beams,       .         .         .         .91 

90.  "  "        "  Steel  Beams, 92 

91.  Neutral  Line — Rectangular  Wrought-Iron  and  Steel  Beams,         .  92 

92.  Transverse  Strength    "  "  "  .  92 

93.  Neutral  Line— Wrt.  Iron  and  St'l  Dbl.  T  and  Rolled  Eye-Beams,  96 

94.  Transverse  Strength  "•«.«•         «•         "         "  96 

95.  Transverse  Strength— Wrought-Iron  and  Steel  Double  T  Beams, 

Flanges  Unequal 96 

96.  Transverse  Strength — Wrought-Iron  and  Steel  Double  T  Beams, 

Flanges  Equal,         .  98 

97.  Neutral  Line — Circular  Wrought-Iron  and  Steel  Beams,       .         .     100 

98.  Transverse  Strength  "  .«..««  .101 


Vlll  CONTENTS. 

ART.  PAGE 

99.  Neutral  Line—  Hollow  Circular  Wrought-Iron  Steel  Beams,        .     104 

100.  Transverse  Strength  "       •"  .         104 

CHAPTER  VI. 

TIMBEIl    BEAMS. 

101.  The  Compress! ve  and  Tensile  Strength  of  Timber,        .         .  .106 

102.  To  Compute  the  Compressive  and  Tensile  Strength  of  Timber,  106 

103.  Neutral  Line— Rectangular  Wooden  Beams,        ....     107 

104.  Transverse  Strength     "                                        ....  108 

105.  Neutral  Line— Circular                "                           .         ...  .111 

106.  Transverse  Strength  "                .  "            "         .         .         .         .  112 

107.  Relative  Strength  of  Square  and  Circular  Timber  Beams,    .  .     113 

CHAPTER  VII. 

STRENGTH    OF    COLUMNS. 

108.  General  Conditions  of  Failure  of  Columns,       .         .         .         .  114 

109.  Resistance  of  the  Cross-Section  of  the  Column,     .        .         .         .119 

110.  Notation, 121 

111.  Deflection  of  Columns, 121 

112.  Classification  of  Columns,           .......  123 

113.  Columns  that  fail  with  the  full  Crushing  Strength  of  the  Material,  125 

114.  Columns  that  fail  with  less  than  the  full  Crushing  Strength  of 

the  Material  and  without  Deflection, 125 

115.  Columns  that  fail  with  less  than  the  full  Crushing  Strength  of  the 

Material  and  with  Deflection,    .         .         .         .         .         .         .127 

116.  Columns  that  Cross-break  from  Compression,  .         .         .         130 

117.  Columns  that  Cross-break  from  Compression  and  Cantileverage,     137 

CHAPTER  VIII. 

COMBINED   BEAMS   AND    COLUMNS. 

118.  General  Statement, 145 

119.  Notation, 146 

INCLINED   BEAMS. 

120.  General  Conditions 147 

121.  Inclined  Beam  Fixed  and  Supported  at  one  end,       .         .         .         148 

CASE  I. —  When  the  Load  is  applied  at  the  free  end  of  the  Beam. 
CASE  II. —  When  the  Load  is  uniformly  distributed  over  the  un- 
supported length  of  the  Beam.  / 


CONTENTS.  IX 

ART.  PAGE 

122.  Inclined  Beam  supported  at  one  end  and  stayed  or  held  in  po- 
sition at  the  other  without  vertical  support,       ....     151 

CASE  I.  —  When  the  Inclined  Beam  is  loaded  at  its  stayed  end. 
CASE  II. —  Wlien  the  Inclined  Beam  is  loaded  at  its  middle. 
CASE  III. —  When  the  Load  is  uniformly  distributed  over  the 

length  of  the  Inclined  Beam. 
CASE  IV. —  When  the  Inclined  Beam  is  loaded  at  its  middle,  and 

with  an  additional  Load  at  its  stayed  end. 
CASE  V. —  When  tlie  Load  is  uniformly  distributed  over  the  length 

of  the  Inclined  Beam,  and  an  additional  Load  applied  to  its 


123.  Inclined  Beam  supported  vertically  at  both  ends  and  loaded  at 

its  middle, 160 

124.  Inclined  Beam  supported  at  both  erids  and  the  Load  uniformly 

distributed  over  its  length, 163 

TRUSSED   BEAMS. 

125.  General  Conditions, 164 

Note. — The  writer  regrets  that  chapters  on  the  following  subjects  are  un- 
avoidably omitted  at  this  time,  "The  Strength  of  Arches"  and  "  The  De- 
flection of  Beams,"  as  they  were  not  satisfactorily  complete,  and  ''Byams 
of  Maximum  Strength  with  Minimum  Material,"  as  he  was  not  fully  pro- 
tected by  letters-patent  at  the  time  of  going  to  press. 


STRENGTH  OF  BEAMS  AND  COLUMNS. 


CHAPTER  I. 

FORCES. 

SECTION  I. — Forces  Defined  and  Classed. 

1.  Force   defined.      Force  may  be   defined  to  be  an 
action  between  two  bodies,  or  parts  of  a  body,  which  either 
causes  or  tends  to  cause  a  change  in  their  condition,  whether 
of  rest  or  of  motion ;  and  our  knowledge  of  a  force  is  prac- 
tically complete   when  we  know    its  intensity,  expressed  in 
pounds  or  some  other  unit  of  measure,  its  direction,  whether 
toward  or   from  the  body   upon   which  it  acts,  its  point  of 
application,  and  the  anyle  that  its  line  of  direction  mak^s 
with  the  surface  of  the  body. 

For  an  unit  of  measure  of  forces  wTe  shall  use  the  pound 
avoirdupois,  as  all  of  our  recorded  knowledge  of  the  strength 
of  materials  used  in  structures  is  expressed  in  pounds  on  the 
square  inch. 

2.  Stress  and  Strain  are  words  used  to  define  that  class 
of  forces  that  are  brought  into  action  when  contiguous  parts 
of  a  body  are  caused  to  react  upon  each  other  by  reason  of 
the  application  of  other  forces  to  the  body,  and  may  be  classed 
as  follows : 

Compression,  Thrust,  or  Pressure  is  the  force  exerted  when 
the  contiguous  parts  of  a  body  are  caused  to  move  toward 
each  other. 

Tension,  Pull,  or  Tensile  Strain  is  the  force  exerted  when 


2  STRENGTH   OF   BEAMS   AND   COLUMNS. 

the  contiguous  parts  of  a  body  are  caused  to  move  away  from 
each  other. 

3.  The  Load.     The  external  forces  applied  to  a  body  to 
produce  the  various  kinds  of    stress  or ' strain  is  called  the 
Load,  and  its  amount  or  magnitude  is  expressed  in  pounds  ;  it 
may  be  classed  as  follows  : 

Concentrated  Forces  or  Loads.  While  in  nature  every  force 
must  be  distributed  over  a  definite  amount  of  surface,  it  is 
necessary,  in  order  to  define  certain  principles,  to  consider 
them  to  be  concentrated  at  a  single  point,  the  effect  being 
identical  with  that  of  the  distributed  load. 

A  Distributed  Force  of  uniform  intensity  is  the  force  or 
load  that  acts  with  the  same  intensity  on  each  square  inch  of 
the  surface  of  the  body  over  which  it  is  distributed. 

A  Distributed  Force  of  uniformly  varying  intensity  is  a 
force  that  increases  in  intensity  in  direct  proportion  to  the 
distance  from  a  given  point. 

4.  Equilibrium   and   Resultant.     Equilibrium  of  a 
system  of  forces  is  such  a  condition  that  the  combined  action 
of  the  forces  produces  no  change  in  the  rest  or  motion  of  the 
body  to  which  it  is  applied. 

The  Resultant  of  a  system  of  forces  is  a  single  concen- 
trated force  that  will  produce  the  same  effect  upon  the  body, 
if  applied,  that  the  system  of  forces  will  produce. 


CONCENTRATED  PARALLEL 


SECTION  II.  —  Concentrated  Parallel  Forces. 

5.  Bending  Moment.  In  this  section  will  be  deduced 
the  relation  existing  between  the  vertical  forces  caused  by 
gravity,  the  supporting  forces  and  the  horizontal  distances 
between  their  lines  of  action.  In  order  to  ascertain  the  rela- 
tion between  such  forces  it  is  necessary  to  obtain  the  product 
of  each  force  by  the  perpendicular  distance  of  its  line  of 
action  from  a  given  point.  Such  product  is  called  the  Bending 
Moment  of  the  force. 

The  vertical  forces  will  be  called  the  loads,  and  the  connec- 
tion between  their  lines  of  action  will  be  called  a  beam,  with- 
out reference  to  the  shape  of  its  cross-section,  its  material,  or 
its  ability  to  resist  the  bending  moment  of  the  applied  loads, 
which  will  be  considered  in  the  sequel. 

The  following  notation  will  be  used  throughout  : 

Z  =  the  total  applied  load  in  pounds. 

,9  =  the  span,  the  horizontal  distance  between  the  supports 
in  inches. 

M  =  the  "bending  moment.  * 

Case  I.  —  THE  BEAM  FIXED  AT  ONE  END  AND  LOADED  AT  THE 
OTHER. 

A  load  thus  applied  will  "bend  the  beam  (Fig.  1)  at  each 


section  from  the  free  end  to  the  point  of  support,  but  un- 
equally.   The  bending  moment  at  any  section  is  the  product  of- 


4         STRENGTH  OF  BEAMS  AND  COLUMNS. 

the  load  by  the  distance  of  the  section  from  the.  free  end  of 
the  beam.  The  Greatest  Bending  Moment,  at  the  point  of 
support,  will  be 

GBM  =  L  ><  *.  (1) 

Case  II. — THE  BEAM  FIXED  AT  ONE  END  AND  THE  LOAD 

UNIFORMLY  DISTRIBUTED  OVER  THE  ENTIRE  SPAN. 

The  fending  moment  at  any  section  is  equal  to  the  product 
of  the  load  between  the  free  end  of  the  beam  and  the  section 
by  one  half  of  its  distance  from  the  free  end ;  the  Greatest 
Bending  Moment,  at  the  point  of  support,  will  be 


or  one  half  what  it  is  in  Case  I.,  the  load,  Z,  and  the  span,  s, 
being  the  same. 

Case  III. — THE  BEAM  SUPPORTED  AT  THE  ENDS  AND  THE 

LOAD  APPLIED  AT  THE  MIDDLE  OF  THE    SPAN. 

In  order  that  equilibrium  shall  exist  one  half  of  the  load 
must  be  supported  by  each  point  of  support ;  the  load  will 


bend  the  beam  (Fig.  2)  in  the  same  manner  that  it  would  if 
the  beam  were  fixed  in  the  middle  of  its  span  and  loaded  at 
each  free  end  with  one  half  of  the  applied  load,  L  ;  the  lend- 
ing moment  at  any  section  will  be  one  half  of  the  load 
multiplied  by  its  distance  from  the  nearest  point  of  support ; 


CONCENTRATED   PARALLEL   FORCES.  5 

the  Greatest  Bending  Moment,  at  the  middle  of  the  span,  will 
be 

Jj         8         -Li  X   S  /o\ 

:  T,  X  77  =  —3 ,  (3) 


or  one  fourth  of  that  in  Case  L,  and  one  half  that  in  Case  II., 
the  load  and  span  being  the  same  in  each. 

Case  IV. — THE  BEAM  SUPPORTED  AT  THE  ENDS  AND  THE 

LOAD,  Z,    UNIFORMLY  DISTRIBUTED  OVER  THE  SPAN. 

The  Greatest  Bending  Moment  occurs  at  the  middle  of  the 
span, 


Case  V. — THE  BEAM  FIXED  AT  BOTH  ENDS  AND  THE  LOAD 

APPLIED  AT  THE  MIDDLE  OF  THE  SPAN. 

In  the  preceding  cases  the  bending  moment  of  the  applied 
load,  Z,  produces  a  strain  of  compression  in  either  the  top  or 
the  bottom  of  the  beam  and  a  tensile  strain  in  the  opposite 
side,  but  in  this  and  the  next  case  the  bending  moment  pro- 


duces  in  the  upper  side,  over  each  point  of  support,  A  A  (Fig.  3), 
a  tensile  strain,  and  in  the  middle  a  compressive  strain ;  and 
in  the  lower  side  at  each  point  of  support,  AA,  a  compressive 
strain,  and  in  the  middle,  m,  a  tensile  strain.  Now,  in  order 
that  these  directly  opposite  strains  may  exist  in  the  upper  and 
lower  sides  of  the  beam  at  the  same  time,  the  strains  at  one 


6         STRENGTH  OP  BEAMS  AND  COLUMNS. 

section  on  each  side  of  the  middle  section  must  be  zero  in 
intensity.  At  these  points,  rr,  the  curvature  of  the  beam 
changes  ;  they  may  be  called  the  points  of  reverse  curvature, 
and  in  this  case  are  located  at  one  fourth  of  the  span  from 
each  point  of  support,  A  A. 

The  Sending  Moments  are  equal  and  greatest  at  three 
sections  ;  at  each  point  of  support  and  at  the  middle  of  the 
span,  theoretically  its  value  is  given  by  the  following  equa- 
tion: 


Barlow  and  other  experimenters  state  that  this  should  be 
GBM=L  *  *. 

D 

Case  VI.—  THE  BEAM  FIXED  AT  BOTH  ENDS  AND  THE  LOAD 

DISTRIBUTED  UNIFORMLY  OVER  THE  SPAN. 

The  sections  at  which  the  Greatest  Bending  Moment  occurs 
are  the  same  as  in  the  preceding  case  ;  but  the  points  of  re- 
verse curvature,  rr,  are  at  the  distance  0.2113s  from  each 
point  of  support,  A  A,  Fig.  3. 

At  the  middle  of  the  span,  m, 


.  (6) 

At  the  points  of  support,  A  A, 

GBM  =  ^p-  (7) 

6.  Case  in  General.  Theory  has  demonstrated  and 
experiments  fully  confirm,  except  as  previously  noted,  that  the 
following  relations  exist  between  the  Greatest  Bending  Mo- 
ments in  beams,  the  span  and  the  total  applied  load  being 


CONCENTRATED   PARALLEL   FORCES.  7 

the  same — that  in  a  beam  fixed  at  one  end  and  loaded  at  the 
other  being  taken  as  our  unit  or  standard  of  measure  : 

n 

Beam  fixed  at  one  end  and  loaded  at  the  other 1 

Beam      "     "     "      "      "         "       uniformly % 

Beam  supported  at  the  ends  and  loaded  at  the  middle J 

Beam         "  "    "       «       "         "      uniformly i 

Beam  fixed  at  both  ends  and  loaded  at  the  middle -J- 

Beam     "      "      «       "       "         "       uniformly TV 

From  which  we  deduce  the  following  formula,  applicable  to 
all  of  the  preceding  cases : 

GSM  =  L  X  *  X  n.  (8) 

Placing  for  the  factor,  n,  the  values  given  in  the  preceding 
table  of  comparison,  the  formulas  heretofore  deduced  for  each 
case  will  be  reproduced. 

These  formulas  for  the  Greatest  Bending  Moments  are 
entirely  independent  of  the  material  composing  the  beam  and 
of  its  cross-section. 


8  STRENGTH   OF   BEAMS   AND   COLUMNS. 


SECTION  III. —  Uniformly  Varying  Forces — Rectangular 

Areas.- 

7.  Notation.     The   principles   governing   the  action  of 
uniformly  varying  forces  distributed  over  rectangular  surfaces 
will  be  deduced,  and  then  those  applicable  to  surfaces  bounded 
by  curved  lines  will  be  considered.     In  order  to  determine  the 
effect  of  such  a  force,  we  must  determine  the  resultant,  its 
point  of  application,  its  lever-arm,  and  from  these  the  moment 
of  the  uniformly  varying  force.     The  following  notation  will 
be  used,  and  it  will  have  the  same  meaning  whenever  it  appears 
in  the  following  pages  : 

C  —  the  maximum  compressive  strain,  in  pounds,  per  square 

inch. 

T  =  the  maximum  tensile  strain,  in  pounds,  pel*  square  inch. 
7?T  =  the  moment  of  the  tensile  strain  in  inch-pounds. 
7?c  =  the        "          "    "   compressive  strain  in  inch-pounds. 
dv  =:  the  depth  of  the  area  covered  by  the  compressive  strain 

in  inches. 
r/T  =  the  depth  of  the  area  covered  by  the  tensile  strain  in 

inches. 

d  =  dc  -\-  dT  =  the  total  depth  of  the  area  in  inches. 
I)  =  the  total  width  of  the  area  in  inches. 

8.  Resultant.     The  amount  of  direct  strain  is  equal  to 

"T 

T 


A 


the  weight  of  a  prismoidal  wedge  composed  of  such  material 
that  a  prism  an  inch  square  in  section  and  C  or  T  high  will 


UNIFORMLY   VARYING  FORCES.  9 

give  a  pressure  on  its  base  equal  to  C  or  T  as  defined  above  ; 
the  resultant  will  then  be  equal  to  the  product  of  the  depth 
r/c  or  d^  by  the  width  5,  by  one  half  of  the  height  C  or  T,  whieli 
is  the  volume  of  the  wedge,  and  it  will  pass  through  a  point 
at  -f  the  depth  dc  or  dT  from  the  edge  of  the  wedge  J.,  of 
which  Fig.  4  is  a  section. 

„      7,     ,      MTT       MCC 
.  • .  Resultant  =  -    -  or  — ^ —  (9) 


Lever-arm  =  —^  or  -^-' 
By  the  Calculus  : 

Let  XT  =  dT  =  AB,  the  depth  of  the  area, 
x  —  any  distance  from  A, 

Tx 
-  =  the  height  of  the  wedge  at  x,  distance  from  A, 

bdx  =  a  small  area,  or  the  differential, 
Ta?  ,  Tbx* 


=  JQ     v"      -  2^' 

.  • .  Resultant  —  — — T> 

/#r  7^2               Tbx3 
bdx  = , 
„      *T                 3^T 


.  *  .  The  moment  —  —  — ?. 
o 

Dividing  the  moment  by  the  resultant  we  obtain  the  lever  arm  =  fdr. 
9.  Problem  I.     REQUIRED  THE  MOMENT  OF  AN  UNIFORMLY 

VARYING    COMPRESSIVE    FOKCE    OF   MAXIMUM   INTENSITY,    C\    WITH 
RESPECT  TO  AN  AXIS,  J3,  IN  THE  BACK  OF  THE  PRESSURE  WEDGE. 

Let  ABD  (Fig.  5)  represent  a  section  through  the  pressure 
wedge. 

MC0 
The  resultant  =  — 7^ — > 


10        STRENGTH  OF  BEAMS  AND  COLUMNS. 

The  lever-arm  =  %dc , 


(11) 


PROBLEM  II.  Required  the  moment  of  only  a  portion  of 
the  force  distributed  over  a  depth  EB  =  dl  and  width  l)l — 
the  force  being  zero  in  intensity  at  A. ;  the  axis  being  at  B 
(Fig.  5).  • 

-•£-7 — !  •  C  =  the  intensity  of  the  force  at  E, 

— '  =  the  lever-arm  of  this  force  at  E,  considered 
2i 

to  be  constant  over  the  distance  EB, 

( 1  —  -£_ — i  J  C  =  the  intensity  of  the  force  at  B  less  that 
atJ?, 

-£  =  its  lever-arm, 
o 

Having  considered  the  force  to  be  divided  into  two  portions, 
the  first,  GE,  constant  in  intensity  over  the  depth  EB, 
and  the  second  increasing  in  intensity  from  zero  at  G  to 

f  1  -      '—j — l  J  G  at  F,  by  adding  the  moments  of  these  parts 
we  obtain 

5  =         (8rf«-H«  ,  (12) 


UNIFORMLY   VARYING   FORCES. 

By  the  Calculus  : 

Let  xc  =  dc  =  the  depth  AB  (Fig.  5), 
x  =  any  distance  from  A, 

— C=  the  intensity  of  the  force  at  the  depth  xt 

bdx  =  a  small  area  of  the  base  of  the  wedge, 


11 


-.JLsBSp.otli.'lt 

Integrating  the  above  expression  for  Rc,  between  the  limits  x  =  dc  and 
x  =  di,  we  obtain 

Ec  —  bjj~  (Me  -  2di)  C,  or  Eq.  12. 
1O.  Problem  III.     REQUIRED  THE  MOMENT,  7?T,  OF  AN 

UNIFORMLY  VARYING  TENSILE  FORCE,  WITH  RESPECT  TO  AN  AXIS, 
0,  PARALLEL  TO  THE  EDGE,  A,  OF  THE  TENSION  WEDGE,  AND  At 
THE  DISTANCE  OA  =  dc  FROM  IT. 

"T 


Let  ABD  (Fig.  6)  represent  a  section  through  the  tension 
wedge. 

-  =  the  resultant, 

a 

df,  -f-  f  dT  =.  its  lever-arm, 


or. 


(13) 


12        STRENGTH  OF  BEAMS  AND  COLUMNS. 

PROBLEM  IV.  Required  the  moment  of  only  a  part  of  the 
above  force  distributed  over  the  depth  EB  =  d2  and  the 
width  ~bv 

Consider  the  force  to  be  divided  into  two  portions.  The 
first,  with  the  intensity  at  E  constant  over  the  distance  EB  ; 

the   second   increasing  in  intensity  from  zero  at   G  to  -j-  T 

«T 
at  F. 

d^  (  1  -    —  -jT  =  the  intensity  of  the  force  at  E^ 
{  d  —  --]  =  its  lever-arm, 

-y  T  =  the  intensity  of  the  force  at  .Z?,  less  that  E^ 

ttj, 

f  6?  —  —  2J  =  its  lever-arm. 
By  adding  the  moments  we  obtain 


~        ~ 

By  the  Calculus  : 

Let  x-t  =  t?T  =  the  distance  AB, 
x  =  any  distance  from  A, 

xT 

—  =  the  intensity  of  the  force  at  any  distance  x, 

(de  -\-x)  =  its  lever-arm, 

bdx  =  the  differential  of  the  area  of  the  base, 


bdr 

.  • .  BT  =  __  (2d  4-  de)  T,  or  Eq.  13. 
o 

Integrating  the  above  expression  for  I&,  between  the  limits  x  =  dr 
and  x  =  di,  and  placing  52  for  b,  we  have 


UNIFORMLY   VARYING   FORCES. 


13 


T. 


This  is  not  identical  in  form  with  Eq.  14,  but  gives  the  same  numerical 
result. 


SECTION  IY. —  Uniformly  Varying  Forces — Circular 
Segment  Areas. 

11.  Resultant  and  Lever-arm.      The  resultant  and 
the  moment  of  an  uniformly  varying  force,  distributed  over 
circular  segment  areas,  cannot  be  readily  deduced  by  means 
of  the  elementary  methods  used  when  the  areas  were  rect- 
angular ;    but  this  can  be  accomplished   with  the  aid  of  the 
Calculus. 

12.  Problem.     REQUIRED  THE  MOMENT  OF  AN  UNIFORMLY 

VARYING  FORCE  DISTRIBUTED  OVER  A  CIRCULAR  SEGMENT,  WITH 
RESPECT  TO  AN  AXIS  THAT  IS  A  TANGENT  TO  THE  BASE  OF  THE 
PRESSURE  WEDGE  AND  PARALLEL  TO  ITS  EDGE,  WHICH  IS  A  CHORD 
OF  THE  CIRCLE. 

Let,  in  Fig.  7,  OBS  represent  the  axis,  MNB  the  base  of 
the  pressure  wedge,  and  ABD  a  section  through  AB. 


r 

c 
k 

3 


O 


Notation. — Adopting    the  notation  defined  in  Art.  7  with 
the  following  in  addition : 


14        STRENGTH  OF  BEAMS  AND  COLUMNS. 

Let  r  —  the  radius  of  the  circle, 
d  =  dc  +  <#T  =  tne  diameter, 
a?c  =  dc  =  the  versine  of  one  half  the  arc  MEN, 
x  =  any  distance  from  the  axis  02?$, 

x*  ~X--  O  —  the  intensity  of  the  pressure  at  any  distance, 

XQ 

x,  from  the  axis  OB8, 
2  (Zrx  —  x^dx  =  the  differential  of  the  area  of  the  base, 

.-.EQ=       X 

i/O 

Integrating,  we  obtain 


Substituting  the  following  equalities  : 


_  I      /V| 

71  versin.  —  ° 


—  one  half  of  the  arc  of  the  segment  MBN 


reducing,  we  obtain 

f1  r~  ~i 

7^c=  24^c     ^«^T  [4rf2c(^-r)  +  r>(30?'-14rfc)]  +Ci?wp.«rc  (12c?e-  lor)»l2    <H& 

from  which  the  moment  of  any  compressed  wedge  may  be 
computed.  The  factor  Comp.  Arc  is  the  arc  MEN. 

13.  Problem.     REQUIRED  THE  MOMENT  OF  AN  UNIFORMLY 

VARYING  TENSILE  FORCE,  DISTRIBUTED  OVER  THE  SEGMENT  OF  A 
CIRCLE,  WITH  RESPECT  TO  AN  AXIS  THAT  IS  A  TANGENT  TO  THE 
CIRCLE,  PARALLEL  TO  THE  EDGE  OF  THE  TENSION  WEDGE,  AND  AT 
THE  DISTANCE  dc  FROM  IT. 

Let,  in  Fig.  8,  FOS  represent  the  axis,  MNB  the  base  of 
the  tension  wedge,  MN  the  edge,  and  OABD  a  section 
through  OB. 


UNIFORMLY   VARYING   FORCES. 


15 


O 


B 


Let  XT  —  dT  —  the  distance  AB, 

x  =  any  distance  from  B  toward  A. 

#T  —  x  j1  —  the  intensity  of  the  force  at  any  distance  a?T  —  a? 

a?T 

from  A., 

d  —  x  •=.  its  lever-arm. 

2  (2rx  —  a?3)*  <&c  =  the  differential  of  the  area  of  the  base  of 

the  wedge. 
For  notation,  refer  to  Arts.  7  and  12. 


=    /Xfe- 


12        24:       24 


9/- 


Substituting  the  following  values  : 


rversin.  —  =  one  half  the  tension  arc 
r 


reducing,  we  obtain 


16 


STRENGTH  OF  BEAMS  AND  COLUMNS. 


Fig.  3 


3?<  -  *•)  +  18r"  ('r  - 
-f-  Tension  arc 


9r)r2)  I  n  6) 


from  which  the  required  moment  may 
be  computed. 

14.  Problem.  REQUIRED  THE  RE- 
SULTANT OR  AMOUNT  OF  DIRECT  FORCE  OF 
AN  UNIFORMLY  VARYING  FORCE  DISTRIB- 
UTED OVER  A  CIRCULAR  SEGMENT  AREA. 

Let    Fig.    9    represent    the    pressure 
wedge,  MBN  the  base  or  circular  seg- 
ment, and  B  the  origin  of  co-ordinates. 
Let  XG  =  AB  the  versine  of  one  half  of  the  arc  MBN, 
x  =  any  distance  from  the  origin  B, 

2  (2rx  —  a?2)  dx  —  the  differential  of  the  segment, 

V  =.  the  resultant  or  volume  of  the  pressure  wedge, 
O=  the  greatest  height,  DB, 


= 

Jo 


Substituting  the  equalities  given  in  Art.  12  and  reducing,  we 
have 


r     r  ~ 


-f-  Segment  arc  (dc  —  r)  3r 


The  resultant  of  a  tension  force  may  be  obtained  from  the 
above  equation  by  placing  T  for  C. 

The  cubic  contents  of  any  cylindrical  wedge  may  be  com- 
puted from  Eq.  17  by  substituting  for  C\  expressed  in  pounds, 
h,  the  greatest  height  of  the  wedge  in  inches. 


UNIFORMLY    VARYING   FORCES. 


17 


The  area  of  the  base  of  the  pressure  wedge  or  that  of  any 
segment  of  a  circle  may  be  computed  from  the  following  : 


Area  =  2  \/dcd^  (^c—  r)  +  Segment  arc  X^-°> 
in  which 

dc  =  versine  of  one  half  the  arc  of  the  segment, 
r  =  the  radius, 
d?  =  the  diameter,  less  dc. 


(VIA) 


SECTION  ^.—  Uniformly  Varying  Forces — Circular 

Arcs. 

15.  Problem.     REQUIRED  THE  MOMENT  OF  AN  UNIFORMLY 

VARYING  COMPRESSIVE  FORCE  DISTRIBUTED  OVER  THE  ARC  OF  A 
CIRCLE,  WITH  RESPECT  TO  AN  AXIS  THAT  IS  A  TANGENT  TO  THE 
CIRCLE  AND  PARALLEL  TO  THE  CHORD  JOINING  THE  EXTREMI- 
TIES OF  THE  ARC. 

Let,  in  Fig.   10,  OBS  represent  the  axis,  MEN  the  arc/ 
ABD  a  section  through  AB,  B  the  origin  of  co-ordinates, 
C  the  intensity  of  the  force  at  B,  and  zero  that  at  M  and  N. 


Fig  10 


For  notation  refer  to  Arts.  7  and  12. 

Let  xc  =  dc  =  the  versine  AB  of  one  half  the  arc 
x  =  any  variable  distance  from  B  or  the  axis, 


18        STRENGTH  OF  BEAMS  AND  COLUMNS. 

^n?    C  =  the  intensity  of  the  force  at  x  distance, 
r  =  the  radius  of  the  circle, 
-  =  twice  the  differential  of  the  arc, 


E  -=  r**  ^rdx       - 

JO  ^%rx   _   g?  X, 

Integrating,  we  obtain 


Jie  =  2C 


r  -  r 


Substituting  the  values  given  in  Art.  12,  we  have 
rV 


(18) 


16.  Problem.    REQUIRED  THE  MOMENT  OF  AN  UNIFORMLY 

VARYING  FORCE  DISTRIBUTED  OVER  THE  ARC  OF  A  CIRCLE,  WITH 
RESPECT  TO  AN  AXIS  THAT  IS  A  TANGENT  TO  THE  CIRCLE,  PARAL- 
LEL TO  THE  CHORD  CONNECTING  THE  EXTREMITIES  OF  THE  ARC, 
AND  AT  THE  DISTANCE  dc  FROM  IT. 

Let,  in  Fig.  11,    FOS  represent   the  axis,  MBN  the  arc, 


T 


OABD  a  section  through  OB,  T  the  intensity  of  the  force  at 
BS  and  zero  at  M  and  N. 


UNIFORMLY   VARYING   FORCES.  19 

For  notation  refer  to  Arts.  7  and  12. 

Let  #T  =  dT  =  the  versine  of  one  half  the  arc  MBN, 
x  =  any  distance  from  the  origin  B, 

g'T  ~  x  T  —  the  intensity  of  the  force  at  any  distance  a?, 

#T 

d  —  x  —  its  lever-arm, 

Zrdx 
.—         —  —  the  differential  of  the  arc. 

V'Zrx  —  X*  ~ 


Integrating,  we  obtain 

B*  =  5  [  4/^  [2  (r  +  4)]  +  Tension  arc  (2//i~ 

from  which  the  required  moment  may  be  computed. 


17.  Problem.     REQUIRED  THE  RESULTANT  OR  AMOUNT  OF 

DIRECT    FORCE    OF    AN     UNIFORMLY  VARYING    FORCE    DISTRIBUTED 
OVER  AN  ARC  OF  A  CIRCLE. 

Let,  in  Fig.  9,  page  16,  MBN  represent  the  arc  of  the  circle, 
1&NBD  a  wedge  whose  cylindrical  surface  is  equal  to  the  re- 
quired resultant,  the  force  being  C  in  intensity  at  B  and  zero 
at  M  and  N. 

Let  xc  —  d  c  =  the  distance  AB, 

x  —  any  distance  from  the  origin  of  co-ordinates  J?, 
a?c  —  x  £,  _  ^  intensity  of  the  force  at  the  point  a?, 


arc> 
V=  the  resultant  or  volume  of  the  pressure  wedge, 

V= 


20  STRENGTH   OF  BEAMS   AND   COLUMNS. 

.'•-.;•  F=  £  fir  VdA  +  Segment  arc  (d0  -  /»)],          (20) 


from  which  the  required  resultant  may  be  computed. 

The  curved  surface  of  a  cylindrical  wedge  may  be  com- 
puted from  the  above  formula  by  substituting  for  (7,  expressed 
in  pounds,  A,  the  greatest  height  of  the  wedge  in  inches. 


CHAPTER  II. 

RESISTANCE  OF  CROSS-SECTIONS  TO   RUPTURE. 

18.  Moment   of   Resistance.       The   cross-section  is 
the  shape  of  the  figure  and  the  area  that  any  material,  such  as 
a  beam,  would  show,  should  it  be  ,cut  into  two  pieces  by  a 
plane  perpendicular  to  its  length,  and  its  resistance  to  rupture 
at  this  plane  or  section  is  the  number  of  inch-pounds  that  its 
fibres  will  offer  to  forces  tending  to  cross-break  the  beam  or 
material  of  which  it  is  a  section  :  this  is  called  the  Moment  of 
Resistance  of  the  cross-section. 

The  Moment  of  Resistance  varies  in  amount  with  the 
material  and  the  shape  of  the  cross-section,  but  it  is  entirely 
independent  of  the  length  of  the  beam  and  of  the  manner  in 
which  the  load  may  be  applied,  in  each  case ;  the  same  cross- 
section  and  material  will  offer  the  same  number  of  inch-bounds 
of  resistance  when  broken  across.  # 

19.  Neutral  Line.     When  a  beam  is  broken  across,  or 
is  acted  upon  by  forces  that  bend  or  tend  to  break  it,  we  know 


from  observation  that  its  fibres  on  the  lower  or  convex  side, 
AB  (Fig.  12),  are  in  a  state  of  tension,  and  that  those  on  the 


22        STRENGTH  OF  BEAMS  AND  COLUMNS. 

upper  or  concave  side,  CD,  are  compressed  ;  but  our  knowl- 
edge obtained  from  observation  is  limited  to  what  takes  place 
oil  the  surface  of  the  beam.  We  can  only  know  what  takes 
place  within  such  a  beam  by  reasoning  from  analogy ;  there  is 
a  tensile  strain  in  the  lower  side  of  the  beam,  AB,  and  just 
the  reverse  character  of  strain  in  the  upper  side,  CD.  In 
order  that  these  two  directly  opposite  strains  may  exist  in  the 
same  beam  at  the  same  time,  both  strains  must  decrease  from 
the  surface  toward  a  common  point  within  the  beam,  where 
both  strains  become  zero  in  intensity,  and  they  may  be  classed 
and  treated  as  'uniformly  varying  forces. 

A  line,  nl,  for  the  longitudinal  section  of  the  beam,  or  a 
plane  for  the  beam,  is  called  the  neutral  line  or  line  of  no 
strain  y  its  position  in  a  beam  having  a  cross-section  of  a 
given  shape,  at  the  instant  of  rupture,  depends  upon  the 
material  alone,  or  upon  the  ratio  existing  between  the  'break- 
ing compressive  and  tensile  fibre  strains.  No  line  in  this  plane 
lias  any  of  the  properties  of  an  axis  that  are  usually  assigned 
to  it  by  writers  on  this  subject. 

2O.  Bending  Moment  and  Moment  of  Resist- 
ance. In  order  to  obtain  the  relation  existing  between  the 
Bending  Moment  of  the  applied  load  and  the  Moment  of 
Resistance  of  the  cross-section  of  the  beam,  conceive  one 
half  of  the  beam,  ABCD  (Fig.  12),  to  be  removed  and  the 
bent-lever,  ogf  (Fig.  13),  to  be  substituted  for  it,  and  that  the 
same  cohesion  to  exist  between  the  fibres  of  the  bent-lever  and 
those  of  the  beam  along  the  line,  fg,  that  originally  existed 
between  the  fibres  of  the  two  halves  of  the  beam  along  the 
same  line,  the  bent-lever,  however,  preserving  its  distinctive 
character  of  a  bent-lever.  The  applied  load,  Z,  causes  the 
bent-lever,  ogf,  and  the  half  of  the  beam,  A  gfD,  to  move 
downward  in  the  direction  of  the  lower  arrow  of  the  figure, 
and  the  end  of  the  lever,  0,  and  that  of  the  half  of  the  beam, 
D,  to  revolve  around  f  in  the  direction  of  the  upper  arrows, 


RESISTANCE   OF   CROSS-SECTIONS   TO   RUPTURE. 


23 


the  effect  being  identical  with  that  produced  by  conceiving 
the  fulcrum,  f,  to  remain  stationary  and  a  moving  force, 
L  -T-  2,  to  be  applied  to  the  end  of  the  bent-lever,  0,  and  at 
the  same  time  to  be  pressed  in  the  direction  of  its  length  by 
a  force,  O,  equal  in  magnitude  to  the  amount  of  direct  com- 
pression along  the  line,y?i,  above  the  neutral  line,  nl.  When 
the  bent-lever  is  made  to  revolve  around  the  fulcrum,/*,  it 
meets  with  an  opposition  of  compression  to  its  motion,  de- 
creasing in  intensity  from  the  fulcrum,  f,  to  the  neutral  ling, 
n,  when  it  becomes  zero  in  value  ;  at  this  point  the  opposition 
changes  to  a  tensile  resistance  which  increases  in  intensity 
from  zero  at  n  to  its  maximum,  T,  at  (j. 

The  bent-lever  and  the  original  vertical  section  of  the 
beam,  fg,  are  pressed  closest  together  at  \he  fulcrum,  f  \  from 
this  point  they  continue  to  separate,  by  virtue  of  the  ductility 
and  compressibility  of  the  material  composing  the  beam,  until 
rupture  takes  place,  eitner  in  its  upper  or  lower  fibres ;  the 
action  of  the  bent-lever  being  identical  with  that  of  the  half 
of  the  beam  for  which  it  was  substituted. 


21,  Equilibrium.    The  Bending  Moment  of  the  applied 
load,  L,  and  the  Moment  of    Resistance  of  the  tensile  and 


24        STRENGTH  OF  BEAMS  AND  COLUMNS. 

compressive  fibre  strains  must  be  taken  or  computed  with 
reference  to  the  fulcrum,  f,  and  in  order  that  equilibrium 
shall  exist,  the  Bending  Moment  of  the  applied  load,  Z,  must 
be  equal  to  the  Moment  of  Resistance,  or  the  sum  of  the 
moments  of  the  tensile  and  compressive  fibre  strains,  and  that 
the  latter  must  be  equal  to  each  other  in  magnitude. 

22.  Position  of  the  Neutral  Line.     By  deducing 
general  formulas  for  the  moments  of  the  tensile  and  compres- 
sive fibre  strain  in  a  cross-section  of  a  given  shape,  placing 
them  equal  to  each  other  and  deducing  from  the  equation  so 
formed  a  general  formula  for  either  gn  or  fn,  the  position  of 
the  neutral  line  may  be  found  in  any  section   of  the  same 
shape  by  substituting  in  the  formula  its  dimensions  and  any 
known  values  for  the  tensile  and  compressive  fibre  strains. 

In  sections  of  an  uniform  shape,  such  as  the  rectangle  and 
the  circle,  the  depth  of  the  neutral  line  below  the  compressed 
side  of  the  beam  may  be  obtained  by  multiplying  the  depth 
of  the  rectangle  and  the  diameter  of  the  circle  by  a  deter- 
mined quantity  that  is  constant  for  each  shape  and  material ; 
this  constant  multiplier  being  its  position  when  the  depth  of 
the  rectangle  and  diameter  of  the  circle  is  unity. 

But  in  irregular  shaped  sections,  such  as  the  T,  Double  T, 
Box,  Rolled-eyebeam  and  other  shapes,  in  which  the  metal  is 
not  continuous  from  the  neutral  line  to  the  top  and  bottom,  a 
special  solution  must  be  made  for  eacli  case  to  determine  the 
position  of  the  neutral  line  from  which  to  compute  the 
Moment  of  Resistance  of  the  section.  Before  beams  of  this 
character  are  manufactured,  an  economical  position  should  be 
assumed  and  a  sufficient  area  of  metal  placed  above  and 
below  the  neutral  line  to  furnish  the  required  Moment  of 
Resistance. 

23.  Neutral    Line    at    the     Transverse    Elastic 

Limit.  Our  object  in  testing,  to  destruction,  the  strength  of 
any  piece  of  construction  material  is  to  obtain  information 


RESISTANCE   OF   CROSS-SECTIONS   TO    RUPTURE.          25 

that  will  guide  us  to  a  correct  knowledge  of  its  use,  when 
safety  to  life  and  property  is  demanded,  whether  this  destruc- 
tion be  by  means  of  extension  or  compression  j  as  we  apply 
the  load  in  small  instalments  there  are  only  two  points  in  its 
intensity  at  which  we  can  record  the  knowledge  thus  gained 
for  future  intelligent,  comparative  use  when  it  has  reached 
the  elastic  limit,  and  when  it  is  sufficiently  intense  to  rupture 
or  destroy  the  piece  of  material  tested.  When  a  beam  is 
broken  by  a  transverse  load  that  has  been  applied  in  small 
instalments,  we  have  two  similar  points,  the  elastic  limit  load 
and  the  rupturing  load,  and  these  are  the  only  points  at 
which  our  information  can  be  used  to  compute  the  strength 
of  similar  material  when  used  in  structures. 

We  know  that  in  an  ordinary  beam  without  a  load  its  com- 
pressive  and  tensile  fibre  strain  is  zero,  and  that  the  breaking 
transverse  load  produces  the  breaking  compressive  and  tensile 
strain  in  the  fibres.  JSTow,  does  the  transverse  elastic  limit 
load  produce  the  elastic  fibre  strain  limits  ?  Is  the  neutral 
surface  the  same  as  that  for  rupture  ?  When  the  beam  is  un- 
loaded each  plane  of  fibres  is  a  neutral  surface  •  as  the  load 
is  applied  the  compressive  and  tensile  strain  penetrates  the 
beam  from  the  top  and  bottom  respectively  ;  theoretically  *we 
know  they  must  meet  at  a  common  point  within  the  beam  at 
the  instant  of  rupture,  and  that  this  is  fully  sustained  by  ex- 
periments will  be  shown  in  the  sequel.  Our  theory  demands 
that  when  the  same  ratio  exists  between  the  tensile  and  com- 
pressive elastic  fibre  strain  limits  that  does  between  the  ulti- 
mate or  breaking  strains,  in  order  that  equilibrium  shall  exist, 
the  neutral  surfaces  must  be  identical,  but  the  theory  does  not 
require  that  the  transverse  elastic  limit  load  shall  produce  the 
elastic  fibre  strain  limits  ;  we  can  only  gain  the  desired  informa- 
tion from  discussing  a  numerical  example. 

The  mean  compressive  and  tensile  elastic  fibre  strain  limits 
for  good  wrought-iron  is  C  —  30000  pounds  and  T  —  30000 
pounds  per  square  inch.  With  these  strains,  from  formulas  de- 


26        STRENGTH  OF  BEAMS  AND  COLUMNS. 

duced  in  the  sequel,  the  centre  elastic  limit  transverse  load  of 
a  bar  of  wrought-iron  six  inches  square  and  ten  feet  span  is 
44100  pounds.  The  centre  elastic  limit  transverse  load  of  a 
bar  of  wrought-iron,  one  inch  square  and  one  foot  span,  is 
2250  pounds,  and  that  of  the  above  beam  from  the  well- 
known  formula  is, 


6  X  36  X  2250 
Load  =  —j—  =  -        ~~i7T~  ~  48600  pounds. 

From  this  practical  identity  of  results,  as  we  have  only  used 
average  values,  and  other  special  tests  given  in  the  sequel,  we 
are  authorized  to  conclude  that  the  transverse  elastic  limit 
load  produces  the  tensile  and  compressive  fibre  strain  elastic 
limits. 

24.  Movement  of  the  Neutral  Line  with  the  De- 
flection. Having  established  the  fact  that  the  transverse 
elastic  limit  load  produces  the  elastic  fibre  strain  limits  —  and 
our  theory  requires  that  the  elastic  limit  neutral  line  and  the 
neutral  line  of  rupture  shall  coincide  only  when  their  ratios 
are  the  same,  but  should  they  be  unequal  they  must  occupy 
different  positions—  in  the  sequel  it  will  be  shown  that  where 
these  ratios  are  unequal  the  elastic  limit  neutral  line  is  situated 
between  the  neutral  line  of  rupture  and  the  bottom  or  ex- 
tended side  of  the  beam,  and  that  as  the  loading  advanced 
from  the  elastic  limit  load  to  the  rupturing  load,  the  neutral 
surface  must  have  moved  upward  or  toward  the  compressed 
surface  of  the  beam. 

From  the  above  we  conclude  that  as  there  was  no  change 
in  the  condition  of  the  loading  that  could  have  reversed  the 
direction  in  which  the  neutral  line  moved  from  its  position  at 
the  elastic  limit  to  that  at  rupture,  the  neutral  line  at  the  in- 
ception of  the  loading  was  at  the  bottom  or  extended  side  of 
the  beam,  and  that  as  the  loading  progressed  it  moved  up- 
ward or  toward  the  compressed  side  of  the  beam—  the  tension 
area,  to  avoid  rupture  in  its  fibres,  continues  to  encroach  upon 


RESISTANCE   OF    CROSS-SECTIONS   TO    RUPTURE.         27 

the  compressed  area  until  the  rupturing  strain  is  produced  in 
both  the  top  and  the  bottom  of  the  beam. 

The  neutral  line,  at  the  inception  of  the  loading,  being  at 
the  bottom  or  extended  side  of  the  beam,  it  can  only  be  moved 
upward  by  reason  of  the  deflection  and  equally  with  it.  If, 
from  the  dimensions  of  the  beam,  it  should  not  be  able  to 
deflect  sufficiently  to  move  the  neutral  line  to  the  position 
required  for  equilibrium  between  moments  of  resistance  of 
the  ultimate  fibre  strains,  the  true  breaking  strength  will  not 
be  obtained  for  the  beam.  When  this  is  the  case  the  observed 
breaking  load  will  be  too  large  for  wooden,  wrought-iron, 
steel  and  tough  cast-iron  beams,  and  too  small  for  the  more 
fractious  varieties  of  material ;  for  should  the  cornpressive 
strain  reach  its  ultimate  limit  before  the  tensile  strain  an  in- 
crease of  the  load  will  develop  a  crushing  strain  in  excess  of 
the  true  crushing  intensity,  as  is  frequently  done  in  crushing 
short  blocks;  the  beam  will,  however,  continue  to  deflect 
under  these  increased  loads,  and  will  finally  develop  the  full 
tensile  strength,  when  the  beam  will  be  broken  by  a  load 
much  in  excess  of  its  true  breaking  load. 

From  the  above,  the  reason  for  the  variation  in  the  modulus 
of  rupture  that  is  required  in  the  "common  theory  ef 
flexure "  is  apparent,  as  the  shorter  beams  in  most  series  of 
experiments,  especially  of  cast-iron,  did  not  deflect  sufficiently 
to  break  with  the  true  breaking  load,  and,  therefore,  it  re- 
quired a  larger  modulus  or  empirical  coefficient  for  the 
shorter  beams  than  for  the  longer  beams  of  the  same  series  of 
tests. 

At  the  instant  of  deflection  the  bending  moment  of  the 
applied  load  is  held  in  equilibrium  by  a  purely  compressive 
resistance,  distributed  over  the  section  as  an  uniformly  vary- 
ing force,  being  zero  in  intensity  at  the  bottom  or  extended 
side  of  the  beam  and  greatest  in  intensity  on  the  opposite  side- 
This  is  a  very  important  principle,  as  from  it  we  shall,  in  the 
sequel,  deduce  the  correct  theory  of  the  strength  of  columns. 


28        STRENGTH  OF  BEAMS  AND  COLUMNS. 

25.  Neutral  Lines  of  Rupture  in  Rectangular 
Sections.     It  may  not  be  out  of  place  here  to  anticipate 
the  results  obtained  in  the  sequel,  by  stating  the  positions  of 
the  neutral  lines  of  rupture  in  a  rectangular  section   when 
composed  of  the  different  kinds  of  material  used  in  construc- 
tion. 

Let  the  section  be  six  inches  deep,  q  the  ratio  of  the  com- 
pressive  to  the  tensile  strains  of  rupture,  or  C  -r-  T. 

c  depth  of  neutral  line  -^ 

below     the    com- 

Cast-iron q  =  8.5      x  ,    . ,      -  , ,      f  2  45  ins 

pressed  side  of  the 

I     section J 

"         q  =  5.  "  "  3.00  " 

"        q  =  4.  "  "  3.24  " 

Steel q  =  1.5  4.29  " 

Wrought-iron  .  . .  .q  =  1.  "  4.68  " 

Beech,  English . .  ..q  =  0.775  «  «  4.90  " 

"         American  q  =  0.383  «  "  5.36  " 

From  this  comparative  statement  we  observe  the  order  in 
which  the  neutral  lines  of  rupture  are  arranged  in  rectangular 
sections ;  the  same  order  of  arrangement  exists  in  all  other 
uniform  sections.  For  different  kinds  of  wood  and  cast-iron 
the  neutral  line  of  rupture  lies  between  the  extremes  given  in 
tlie  above  table  of  comparison. 

26.  Relative   Value  of  the  Compressive  and 
Tensile  Strains.      Experiments  have    fully  shown    that 
the  compressive  and  tensile  strains  do  not  possess  equal  values 
as  factors  in  determining  the  transverse  load  that  a  beam  will 
bear,  and  that  the   influence  of  the  tensile  strain   predomi- 
nates. 

The  elastic  limit  or  technical  breaking  load  of  a  wrought- 
iron  beam  one  inch  square  and  twelve  inches  span,  loaded  in 
the  middle,  is  2000  pounds  when 


RESISTANCE    OF    CROSS-SECTIONS   TO    RUPTURE.          29 

C  =  30000,  T  =  30000  pounds,  and  the  Moment  of  Re- 
sistance  =  6084  inch-pounds  from  our  formulas. 

We  will  now  endeavor  to  trace  the  effect  that  will  be  pro- 
duced upon  the  amount  of  the  transverse  breaking  load  from 
varying  the  exact  and  relative  values  of  C  and  T  from  those 
in  the  above-described  wrought-iron  beam,  which  will  be  taken 
as  our  standard  of  comparison.  Since  the  crushing  and 
tensile  strength  of  the  material  composing  any  beam  must 
each  sustain  one  half  of  the  breaking  load  of  the  beam,  the 
per  cent  of  loss  or  gain  in  the  transverse  strength  should  be, 
approximately,  one  half  the  sum  of  the  per  cents  of  the  losses 
or  gains  in  the  values  of  C  and  T,  but  this  will  be  found  to 
be  true  for  only  the  smaller  ratios  of  C  -T-  T  that  exist  in 
materials  of  construction. 

The  centre  breaking  load  of  a  white  pine  beam  one  inch 
square  and  twelve  inches  span  is  450  pounds  when 

C  =  5000,  T  =  10000  pounds,  and  the  Moment  of  Re- 
sistance  =  1260  inch-pounds.  In  passing  from  the  standard 
wrought-iron  beam  where  C  =  T  to  the  white  pine  beam 
where  C  —  0.5  T,  the  following  changes  take  place  : 

Loss  in  the  value  of  C 83.4  per  cent, 

«  "  T. 66.6         " 

Apparent  loss  to  the  transverse  load 75.0 

Actual         «       «  "  77.5 

Loss  to  the  Moment  of  Resistance 79.2 

results,  practically,  identical  for  this  ratio. 

From  Mr.  Kirkaldy's  experiments  a  bar  of  steel  one  inch 
square  and  twelve  inches  span  will  break  with  a  centre  load 
of  6400  pounds  when 

C  —  160000,  T  =  70000  pounds,  Moment  of  Resistance 
=  19200  inch-pounds.  In  passing  from  our  standard  wrought- 
iron  beam  to  the  steel  beam,  the  following  changes  take 
place  : 


30        STRENGTH  OF  BEAMS  AND  COLUMNS. 

Gain  in  the  value  of  C 433.3  per  cent. 

"  "  T 133.3         " 

•  

Apparent  gain  to  the  transverse  load 283.3         " 

Actual          "        "  "  "    220.0         " 

Gain  to  the  Moment  of  Resistance 215.0 

Mr.  Hodgkinson  found  the  centre  breaking  load  of  a  certain 
cast-iron  beam,  one  inch  square  and  twelve  inches  span,  to  be 
2000  pounds  when 

C  =  115000,  T  =  14200  pounds,  and  the  Moment  of  Re- 
sistance =  6600  inch-pounds.  In  passing  from  our  standard 
wrought-iron  to  the  cast-iron  beam  of  the  same  size,  the 
following  changes  take  place  : 

Gain  in  the  value  of  C -|-  283.4  per  cent. 

Loss   "     "       "      "    T . .  .    —    52.6    "      " 


Apparent  gain  to  the  transverse  load  ....  115.4    "      " 

Actual          "        "  "  "      0.0    "      " 

Gain  to  the  Moment  of  Resistance 8.4    "      " 

In  this  experiment  it  required  283.4  per  cent  gain  in  the 
value  of  C  to  offset  a  loss  of  52.6  per  cent  in  the  value  of  T, 
or  that  the  compressive  strength  does  not  sustain  its  proper 
proportion  of  the  load. 

This  great  discrepancy  between  the  legitimate  theoretical 
deductions  and  the  results  obtained  from  experiments  cannot 
be  reconciled  on  the  hypothesis  that  the  forces  are  in  equi- 
librium with  respect  to  an  axis  that  lies  writhin  the  beam, 
the  moment  in  each  case,  for  rectangular  sections,  being  the 
resultants  of  the  tensile  and  compressive  forces,  multiplied  by 
two  thirds  of  the  depth  of  their  respective  areas,  showing  that 
the  compressive  strain  works  under  no  disadvantages  ;  but  on 
our  theory  this  discrepancy  is  fully  accounted  for.  The  lever- 
arm  of  the  crushing  resultant  is  one  third  the  depth  of  the 


RESISTANCE   OF    CROSS-SECTIONS   TO   KUPTUKE  31 

compressed   area,   while  that  of  the  tensile  resultant  is   only 
one  third  less  than  the  total  depth  of  rectangular  beams. 

27.  Summary  of  the  Theory.  The  theory  herein 
advanced  to  explain  the  relation  that  exists  between  the  Bend- 
ing Moment  of  the  applied  load  and  the  Moment  of  Resist- 
ance of  the  material  composing  the  beam,  may  be  expressed  by 
the  following  hypotheses  : 

1st.  The  fibres  of  the  beam  on  its  convex  side  are  extended 
and  those  on  the  concave  side  are  compressed  in  the  direction 
of  the  length  of  the  beam,  and  there  are  no  strains  but  those 
of  extension  and  compression. 

2d.  There  is  a  layer  or  plane  of  fibres  between  the  extended 
1  and  compressed  sides  of  the  beam  that  is  neither  extended  nor 
compressed,  which  is  called  the  neutral  surface  or  neutral  line 
for  any  line  in  this  plane. 

3d.  The  strains  of  compression  and  extension  in  the  fibres  of 
the  beam  are,  in  intensity,  directly  proportional  to  their  dis- 
tance from  the  neutral  surface. 

4th.  The  axis  or  origin  of  moments  for  the  tensile  and  com- 
pressive  resistance  of  the  fibres  of  any  section  at  right  angles 
to  length  of  the  beam,  is  a  line  of  the  section  at  its  intersection 
with  the  top  or  compressed  side  of  the  beam. 

5th.  The  fibres  of  a  beam  will  be  ruptured  by  either  the 
tensile  or  compressive  strains  in  its  concave  and  convex  sur- 
faces, whenever  they  reach  in  intensity  those  found  by  experi- 
ment to  be  the  direct  breaking  tensile  and  compressive  fibre 
strains  for  the  material  composing  the  beam. 

6th.  The  Bending  Moment  of  the  load  at  any  section  is 
equal  to  the  sum  of  the  moments  of  resistance  to  compression 
and  extension  of  the  fibres,  or  to  the  Moment  of  Resistance  of 
the  section  of  the  beam. 

7th.  The  sum  of  the  moments  of  resistance  of  the  fibres  to 
compression  is  equal  to  the  sum  of  the  moments  of  resistance 
of  the  fibres  to  extension. 


32  STRENGTH   OP   BEAMS   AND    COLUMNS. 

8th.  The  algebraic  sum  of  the  direct  forces  of  compression 
and  extension  can  never  become  zero. 

9th.  The  Moment  of  Resistance  of  the  section  is  equal  to 
the  sum  of  the  moments  of  resistance  to  the  compression  and 
extension  of  its  fibres. 

10th.  The  transverse  elastic  limit  load  produces  the  tensile 
and  compressive  fibre  strain  elastic  limits. 


CHAPTER  III. 

TRANSVERSE  STRENGTH. 
SECTION  I. —  General    Conditions. 

28.  Coefficients  or  Moduli  of  Strength  are  quanti- 
ties expressing  the  intensity  of  the  strain  under  which  a  piece 
of   a    given   material  gives  way  when   strained   in   a  given 
manner,  such  intensity  being  expressed  in  units  of  weight  for 
each  unit  of  sectional  area  of   the  material  over  which  the 
strain  is  distributed.     The  unit  of  weight  ordinarily  employed 
in  expressing  the  strength  of  materials  is  the  number  of  pounds 
avoirdupois  on  the  square  inch. 

Coefficients  of  Strength  are  of  as  many  different  kinds  as 
there  are  different  ways  of  breaking  a  piece  of  material. 

•Coefficients  of  Tensile  Strength  or  Tenacity  is  the  strain 
necessary  to  rupture  or  pull  apart  a  prismatic  bar  of  any  given 
material  whose  section  is  one  square  inch,  when  acting  in  the 
direction  of  the  length  of  the  bar.  This  strain  is  the  T  of 
our  formulas. 

Coefficient  of  Crushing  Strength  or  Compression  is  the 
pressure  required  to  crush  a  prism  of  a  given  material  whose 
section  is  one  square  inch,  and  whose  length  does  not  exceed 
from  one  iofive  times  its  diameter,  in  order  that  there  may  be 
no  tendency  to  give  way  by  bending  sideways.  This  pressure 
is  the  C  of  our  formulas. 

29.  Elasticity  of  Materials.     It  is  found  by  experi- 
ment that  if  the  load  necessary  to  produce  a  strain  and  fracture 
of  a  given  kind  is  applied  in  small  instalments,  that  before 
the  load  becomes  sufficiently  intense  to  produce  rupture,  it  will 


34  STRENGTH   OF   BEAMS   AND    COLUMNS. 

cause  a  change  to  take  place  in  the  form  of  the  material,  and 
if  the  load  is  removed  before  this  intensity  of  the  fibre  strain 
passes  certain  limits,  the  material  possesses  the  power  of  re- 
turning to  its  original  form.  This  is  called  its  elasticity. 

3D.  Elastic  Limits.  When  the  material  possesses  the 
power  of  recovering  its  exact  original  form  without  "  set"  on 
the  removal  of  a  load  of  a  given  intensity,  the  greatest  load 
under  which  it  will  do  this  is  called  the  limit  of  perfect  elas- 
ticity. 

The  limit  of  elasticity  as  ordinarily  defined  and  used  by 
experimenters  is  that  point  or  intensity  of  strain  where  equal 
instalments  or  increments  of  the  applied  load  cease  to  pro- 
duce equal  changes  of  form,  or  where  the  change  in  form  in- 
creases more  rapidly  than  the  load. 

31.  The   Elastic   Limit  of  Beams  may   be   deter- 
mined by  applying  small  equal  parts  of  the  load  and  noting 
the  increase  in  deflection  after  each  increase  of  the  load,  al- 
lowing  sufficient   time  for  each  increase  of  the  load  to  pro- 
duce its  full  effect.     "When  it  is  found  that  the  deflections 
increase  more  rapidly  than  the  load,  its  elastic  limit  has  been 
reached  and  passed.     The  relation  between   the   elastic  limit 
load  of  the  beam  and  the  elastic  limit  of  the  tensile  and  com- 
pressive  fibre  strains  of  the  material  composing  the  beam  will 
be  shown  in  the  sequel,  or  that  the  elastic  limit  load  of  the 
beam  produces  the  elastic  limit  strain  for  the  fibres. 

32.  Working    Load    and    Factor    of    Safety. 

The  greatest  load  that  any  piece  of  material,  used  in  a  struc- 
ture, is  expected  to  bear  is  called  the  working  load. 

The  'breaking  load  to  be  provided  for  in  designing  a  piece 
of  material  to  be  used  in  a  structure  is  made  greater  than  the 
working  load  in  a  certain  ratio  that  is  determined  from  ex- 
perience, in  order  to  provide  for  unforeseen  defects  in  the 
material  and  a  possible  increase  in  the  magnitude  of  the  ex- 
pected working  load. 


TRANSVERSE   STRENGTH.  85 

The  factor  of  safety  is  the  ratio  or  quotient  obtained  by 
dividing  the  breaking  load  by  the  working  load  required. 

33.  General  Formula.       In    our    first  chapter  we 
deduced  rules  or  formulas,  from  which  can  be  computed  the 
Greatest  Bending  Moment  that  a  load  applied  to  a  beam  in  a 
given  manner  will  produce  without  reference  to  the  shape  of 
its  cross-section,  or  to  the  material  composing  the  beam. 

In  our  second  chapter  principles  are  deduced  from  which 
can  be  computed  the  Greatest  Moment  of  Resistance  cross- 
sections  of  the  various  shapes  and  material  will  exert  at  the 
instant  of  rupture,  without  reference  to  the  length  of  the  beam 
or  to  the  manner  in  which  the  load  may  be  applied. 

To  avoid  repetition,  the  formulas  for  the  Moments  of  Re- 
sistance are  deduced  in  this  chapter.  Our  knowledge  of  the 
transverse  strength  of  beams  will  now  be  complete  if  we  com- 
pute and  make  the  Greatest  Moment  of  Resistance  of  the 
cross-section  of  the  beam  equal  to  the  Greatest  Bending 
Moment  of  the  applied  load. 

Let  1?  =  the  Greatest  Moment  of  Resistance  of  the  'beam, 
L  =  the  total  applied  load  in  pounds,  , 

s  =  the  span,  the  distance  between  the  supports  in 

inches, 

n  =  the  factor  defined  in  Art.  6,  page  7, 
M  =  the   Greatest    Bending   Moment' of   the  applied 
load. 

From  Eq.  8,  page  7,  we  have 

M  =  nLs  —  R, 

.  •  .     L  =  -,  (21) 

ns 

from  which  the  breaking  load  of  the  beam  may  be  computed. 

34.  Relative  Transverse  Strength  of  a  Beam. 

Referring  to  the  values  of  the  factor,  n,  of  Eq.  21,  as  given 


36        STRENGTH  OF  BEAMS  AND  COLUMNS. 

in  Art.  6,  page  7,  it  is  found  in  each  case  to  be  either 
unity  or  a  fraction,  and  when  these  values  are  introduced  for 
7i,  in  Eq.  21,  it  is  equivalent  to  multiplying  the  numerator,  7?, 
by  the  denominator  of  the  fraction,  n ;  hence  if  we  make 

1 

n  ~ 
Eq.  21  will  become 

L  =  ^.  (22) 

Computing  and  tabulating  the  values  of  m  from  those  of  n, 
Art.  6,  we  obtain  the  following  relation  between  the  breaking 
loads  of  a  beam,  the  span,  material  and  cross-section  remain- 
ing the  same  or  constant ;  the  breaking  load  of  a  beam,  fixed 
at  one  end  and  loaded  at  the  other,  being  the  unit  or  standard 
of  strength. 

m 

Beam  fixed  at  one  end  and  loaded  at  the  other 1 

Beam     "  "        "        "       uniformly 2 

Beam  supported  at  its  ends  and  loaded  at  its  middle 4 

Beam  "  "         "       "        "       uniformly 8 

Beam  fixed  at  both  ends  and  loaded  at  its  middle 8 

Beam       «       "  «        «         «      uniformly 12 

Eq.  22  is  the  general  formula  from  which  will  be  de- 
duced the  transverse  strength  of  beams  of  all  sections  by 
giving  to  the  factor,  7?,  its  proper  value. 


TRANSVERSE   STRENGTH. 


37 


SECTION  II. — Transverse  Strength — Rectangular  Sections. 

35.  Moment  of  Resistance.  In  Fig.  14,  let  AEDX 
represent  .the  section,  nl  the  neutral  line,  AnlX  the  compressed 
area,  BnlD  the  extended  area,  AnBMnNA  any  section  through 


c 
i    4. 


the  strain  wedges  at  right  angles  to  the  neutral  line,  and  AX 
the  axis  or  origin.  Adopting  the  notation  heretofore  used  in 
Art.  7,  we  have 

C  —  the  greatest   intensity,  per  square  inch,  in  peunds   of 
the  compressive  strain, 

T  =  the    greatest   intensity,  per  square  inch,  in  pounds  tff 
the  tensile  strain, 

7?c  =  the  moment  of  the  compressive  strain  in  inch-pounds, 

7?T  =  the  moment  of  the  tensile  strain  in  inch-pounds, 

R  =  T^c  +  7?T  =  the  Moment  of  Resistance  of  the  section  in 

inch  pounds, 

dc  —  the  depth  of  the  compressed  area,  An,  in  inches, 
r/T  =  the  depth  of  the  extended  area,  Bn,  in  inches, 
d  =  dc  -\-  dT  =  the  depth  of  the  section  AB  in  inches, 
J   =  the  width  of  the  section  BD  in  inches, 
q   =2  the  quotient  arising  from  dividing  O  by  T7, 

L   =  the  total  applied  load  in  pounds, 
s  =  the  span,  the  distance  between  the  supports  in  inches, 

m  =  the  factor,  defined  in  Art.  34,  page  35. 


38        STRENGTH  OF  BEAMS  AND  COLUMNS. 

From  Eq.  11,  page  10,  we  have,  for  the  moment  of  com- 
pressive  resistance, 

£c  =  W^L,  (23) 

and  for  the  moment  of  the  tensile  resistance  from  Eq.  13, 

je,  =  M(2d+^)7.  (24) 

D 

36.  The  Neutral  Line.  From  the  7th  hypothesis,  Art. 
23,  page  31,  we  have,  from  equating  the  second  members  of 
Eqs.  23  and  24, 


6 
from  which 


q+l 
and 

^(-0.5+1/27+2725) 

"   £  +  1" 

The  position  of  the  neutral  line  in  any  rectangular  sec- 
tion may  be  computed  from  either  of  these  formulas  ;  it  is 
independent  of  the  absolute  intensity  of  the  maximum  tensile 
and  compressive  fibre  strains,  but  depends  upon  their  ratio 
C-T-  T=  q  for  its  position.  The  application  of  these  formulas 
is  illustrated  in  Examples  5,  6  and  7  of  the  sequel. 

37,  Transverse  Strength,  General  formulas  for  the 
transverse  strength  may  be  obtained  by  substituting  for  JR  the 
Moment  of  Kesistance  in  Eq.  22,  page  36,  its  values  27?c  —  2#T 
from  Eqs.  23  and  24. 


(27) 
.  L  =  (2d+  &)  T.  (28) 


TRANSVERSE   STRENGTH.  39 

From  either  of  these  equations,  either  the  breaking  load  or 
the  elastic  limit  load  may  be  computed  by  giving  to  C  and  1 
the  breaking  or  elastic  limit  values  of  the  material  composing 
the  beam.  The  application  of  Eq.  27  is  illustrated  in  Exam- 
ples 7,  26,  27,  28,  36,  37,  38,  39  and  40,  and  Eq.  28  in  Ex- 
amples 5,  6,  7,  8,  25  and  30  of  the  sequel. 

38.  To  Design  a  Beam.     In  designing  a  beam  of  a 
rectangular  section  that  shall  break  with  a  given  total  applied 
load,  distributed  over  the  span  of  the  beam  in  a  given  manner, 
the  span  of  the  beam  will  be  determined  from  the  position  in 
which  it  is  to  be  used,  and  the  depth,  d,  in  inches  will  be  as- 
sumed ;  the  crushing  and  tensile  strength  of    the   material 
composing  the  beam  must  be  known  •  it  will  then  only  be  nec- 
essary to  determine  the  width,  6,  in  order  that  the  beam  shall 
break  with  the  required  load. 

From  Eq.  25  deduce  the  position  of  the  neutral  line,  which 
is  independent  of  the  width,  £,  then  from  Eq.  27  deduce  the 
value  of  &,  the  width 

(29) 

which  gives  the  required  width,  since  all  of  the  factors  in  the 
second  member  of  this  equation  are  known  quantities. 

Should  the  assumed  depth  and  the  computed  width  not  give 
an  economical  section  for  the  beam,  a  second  depth  must  be  as- 
sumed from  this  information  and  a  second  width  computed ; 
this  process  should  be  repeated  until  a  satisfactory  result  is 
secured. 

39.  To  Compute  the  Compressive  and  Tensile 
Strains.    , 

PROBLEM  I. — The  position  of  the  neutral  line  and  the 
crushing  strain  of  the  material  of  a  rectangular  section  may 
be  computed  from  the  known  transverse  breaking  load  and 
tensile  strength. 


40        STRENGTH  OF  BEAMS  AND  COLUMNS. 

From  Eq.  28,  page  38,  deduce  the  value  of  r/T, 


7  _  m 

:T~  " 

and 


which  gives  the  required  position   of  the  neutral  line,  illus- 
trated in  Examples  1,  2,  21,  22  and  35. 

The  crushing  strength  can  be  computed  by  deducing  its 
value,  6",  from  Eq.  27, 

.•.<7=J^L,  (31) 

mbdc 

The  application  is  illustrated  in  Examples  1,  2  and  21. 

PROBLEM  II.  —  The  position  of  the  neutral  line  and  the 
tensile  strength  of  the  material  of  a  rectangular  section  may 
be  computed  from  the  known  transverse  breaking  load  of  the 
beam  and  the  crushing  strength  of  the  material. 

From  Eq.  27  we  have 

(32) 


for  the  position  of  the  neutral  line,  which  is  illustrated  in  Ex- 
ample 4. 

From  Eq.  28  we  have  for  the  required  tensile  strength 


=          ~'  (33) 

illustrated  in  Example  4. 


TRANSVERSE   STRENGTH. 


41 


SECTION  III.  —  Transverse  Strength — Hodgkinson  Section. 

4O.  Moment  of  Resistance.    In  Fig.  15  let  ABDX 
represent  the  section,  A  OX  the  axis  or  origin  of  moments, 


nl  the  neutral  line,  the  area  above  it  being  compressed  and 
that  below  it  extended. 

OnEMnNO,  a  section  through  the  strain  wedges  on  the 
line  OE. 

d  =  the  depth  of  the  web  and  section,  OE,  in  inches, 

dl  =  the       "         "       upper  flanges  in  inches, 

dt  =  the       "         "       lower         "  " 

1)   —  the  width  of  the  web  in  inches, 

Jj  —  the  sum  of  the  widths  of  the  upper  flanges  in  inches, 

or  AX  —  &, 
&2  —  the  sum  of  the  widths  of  the  lower  flanges  in  inches, 

or  BD  -  I. 

For  other  notation  refer  to  Art.  35. 

From  Eqs.  11  and  12,  page  10,  we  have  for  the  moments  of 
eompressive  resistance, 

ld*C 
Web,  7?c  =  —  -, 


Upper  flanges,  7?c  =        -  (3d  -  2d>)  C. 


42  STRENGTH   OF   BEAMS   AND   COLUMNS. 

Adding,  we  obtain  for  the  section, 


and  for  the  moments  of  tensile  resistance  from  Eqs.  13  and 
14,  page  11, 


Lower  flanges,  ^T  =  gy  [zdtdT  (2^  -  dt)  —  d?  (3d  -  M,)~\  T. 

Adding  the  above  equations  we  obtain  for  the  moment  of 
tensile  resistance  of  the  section, 


R,=    WTW+rfc)  +!,,[3dA(2d-d,)-d,'(3d-2d,)]          (35) 


and  the  Moment  of  Resistance  of  the  Hodgkinson  section  is 
=  2/4  from  Eqs.  34  and  35. 


41.  The  Neutral  Line.  By  equating  the  values  of 
ftc  and  7?T  given  by  Eqs.  34  and  35,  and  deducing  from  the 
equation  so  formed  the  value  of  d^  the  position  of  the  neutral 
line  is  determined,  but  as  this  will  involve  the  solution  of  a 
biquadratic  equation,  the  general  solution  will  be  too  complex 
for  ordinary  use. 

The  following  approximate  formula  obtained  by  neglecting 
certain  quantities  that  do  not  materially  affect  the  result  in 
ordinary  cases,  will  give  its  position  sufficiently  near  for  all 
practical  purposes  when  the  beam  is  made  of  cast-iron  : 


d  =  d  +  ^m<MA  (q 


(q  -f   1) 

Its  application  is  illustrated  in  Examples  8,  9,  10  and  29. 

The  position  of  the  neutral  line  in  .a  Hodgkinson  or  in 
any  single  or  double  flanged  beam  may  be  computed  when 
the  transverse  strength  of  the  flanged  beam  and  either  the 


TRANSVERSE   STRENGTH.  43 

« 

compress! ve  or  tensile  strength  of  the  material  have  been  deter- 
mined experimentally. 

PROBLEM  I. — Required  the  position  of  the  neutral  line  in 
any  flanged  beam  when  tlw  transverse,  ultimate,  or  elastic 
load  and  the  corresponding  compressive  strength  of  the 
material  have  been  obtained  from  experiments. 

For  any  given  flanged  beam  with  its  transverse  and  com- 
pressive strength  determined  experimentally,  all  of  the  terms  of 
Eq  37,  page  44,  except  d^  become  known  quantities ;  by 
giving  the  letters  their  numerical  values  for  any  beam,  the 
formula  can  always  be  reduced  to  the  following  general  form : 

d<?  ±  3/*4  ±  2&  —  0, 

in  which  3p  and  2&  are  numerical  quantities ;  then  by  Car- 
dan's Rule  for  the  solution  of  cubic  equations  of  this  form, 


dc  =       k  +   |/£>  +y  +      &  _     !/#  _|_  p>9      (A  36) 

p  and  Jc  must  be  made  equal  to  one  third  and  one  half  of 
the  numerical  quantities,  Zp  and  2&,  respectively,  and  their 
algebraic  sign  must  be  that  of  the  term  in  which  they  appear 
in  the  reduced  equation. 

From  the  above  equation  one  value  of  dc  will  be  determined, 
and  with  it,  Eq.  (A  36)  may  be  reduced  to  an  equation  of  the 
second  degree  by  dividing  the  cubic  equation  by  dc  plus  or 
minus  the  numerical  value  deduced  above,  giving  the  numer- 
ical value  in  the  division  the  reverse  sign  to  that  computed. 
From  this  resulting  equation  of  the  second  degree,  the  other 
two  values  of  dQ  may  be  computed ;  the  value  that  represents 
the  position  of  the  neutral  line  may  be  generally  determined 
from  inspection. 

PROBLEM  II. — Required  the  position  of  the  neutral  line  in 
any  flanged  beam  when  the  transverse  load  and  tensile 
strength  of  the  material  have  been  obtained  from  experiments. 

In  Eq.  38,  which  gives  the  relation  between  the  Moment  of 


44        STRENGTH  OF  BEAMS  AND  COLUMNS. 

the  applied  load  and  the  Moment  of  Resistance  of  the  section, 
all  of  the  terms  become  known  quantities  except  dT  ;  substitut- 
ing for  the  letters  their  values  in  any  given  case,  it  will 
always  reduce  to  the  following  general  form  : 

dj  —  adT*  ±  mdT  ±  n  —  0, 

in  which  $,  w,  and  n  are  numerical  quantities.  By  substitut- 
ing for  dT  a  new  unknown  quantity, 


the  equation  will  reduce  to  the  following  general  form  : 

x3  ±  Zpx  ±  2&  —  0,  (B  36) 

which  can  be  solved  by  Cardan's  Rule,  as  in  the  preceding 
Problem. 

42.  Transverse  Strength.  To  obtain  a  general 
formula  from  which  the  transverse  strength  of  a  Hodgkinson 
beam  may  be  computed,  place  for  7?,  the  Moment  of  Resist- 
ance in  Eq.  22,  page  36,  its  values,  2#c  =  2^x,  from  Eqs.  34 
and  35,  and  we  have 


L  =  H1  +  M,1  (3rfc  -  3d,),  (37) 

and 


From  either  of  these  formulas,  either  the  breaking  or  elastic 
limit  load  may  be  computed.  Their  application  is  illustrated 
in  Examples  8,  9,  10,  22,  29,  30  and  31. 

43.  To  Design  a  Hodgkinson  Beam.  An  econom- 
ical position  for  the  neutral  line  must  be  assumed  and  a  suffi- 
cient area  of  the  metal  arranged  above  and  below  it,  to  furnish 
the  required  compressive  and  tensile  resistance  ;  the  depth  of 


TRANSVERSE   STRENGTH.  45 

the  beam,  d,  and  the  thickness  of  the  web,  ?>,  should  also  be 
assumed. 

Let  L  =  the  required  breaking  load  in  pounds, 

s  ='the  span  in  inches, 
m  =  the  factor  defined  in  Art.  35. 

Step  I.  —  Assume  the  thickness  of  the  web,  &,  and  its 
depth,  d,  which  is  also  the  depth  of  the  beam,  and  an  economi- 
cal position  for  the  neutral  line.  The  compressed  area  above 
the  neutral  line  must  furnish  one  half  and  the  extended  area 
below  it  the  other  half  of  the  required  strength  of  the  beam. 

FOR    THE    COMPRESSED    AREA  I 

Step  II.  —  Place  the  value  of  the  compressive  resistance 
given  by  Eq.  11,  page  10,  for  7?  in  Eq.  22,  page  36,  and  we 
shall  have  for  the  load,  Z,,  sustained  by  the  compressed  area  of 

the  web, 

mbd*C 


Step  III.  —  For  the  top  flanges  deduct  the  load,  Z15  ob- 
tained in  Step  II.  ,  from  one  half  of  the  required  breaking 
load,  Z,  of  the  beam  ;  the  balance  is  the  breaking  load,  Z2,  for 
the  top  flanges,  or, 


In  order  to  design  an  area  of  section  that  will  sustain  this 
load,  we  must  assume  a  convenient  depth,  d^  for  the  top 
flanges  and  obtain  the  sum  of  their  widths,  J,,  by  placing  for 
7?,  in  Eq.  22,  page  36,  its  value  for  this  case,  from  Eq.  12,  page 
10,  and  we  will  have 

_  W  (3dc  -  2rf.)  G  , 

L"  -         ~W 

from  which  we  have 

7       __  §dcL 
~ 


46        STRENGTH  OF  BEAMS  AND  COLUMNS. 

One  half  of  this  computed  width,  J,,  must  be  arranged  on 
each  side  of  the  web. 

FOR   THE    EXTENDED    AREA  : 

Step  IV,—  Substitute  for  R,  in  Eq.  22,  page  36,  its  value 
from  Eq.  13,  page  11,  and  we  will  have  for  the  load,  Z3,  sus- 
tained by  the  extended  area  of  the  web,  * 

(43) 

Step  V,  —  For  the  bottom  flanges  deduct  the  load,  Zs, 
found  by  Step  IY.,  from  one  half  of  the  required  breaking 
load,  Z,  of  the  beam  ;  the  remainder  will  be  the  load,  Z4,  that 
must  be  sustained  by  the  bottom  flanges,  or 

Lt  =  ^-Ly  '       (44) 

In  order  to  design  an  area  for  the  bottom  flanges  that  will 
sustain  this  load  Z4,  we  must  assume  a  convenient  depth,  d» 
for  the  bottom  flanges,  and  obtain  the  sum  of  their  widths,  J2, 
by  substituting  for  R,  in  Eq.  22,  its  value  for  this  case  from 
Eq.  14,  page  12,  and  we  will  have 


?,/*  (2d  -  4)  -  d;  (3d  -  2^)  ,        (45) 

from  which  we  deduce 


____ 

-  d,)  -  d?  (3d  -  2<72)]  mT 


One   half  the   width   computed  from  this  formula  must  be 
arranged  on  each  side  of  the  web. 

Step  VI.  —  Should  the  computed  dimensions  from  those 
assumed  produce  a  badly  designed  section,  new  dimensions 
must  be  assumed  from  the  knowledge  thus  obtained,  and  a 
second  computation  made  as  in  the  first  instance. 


TRANSVERSE   STRENGTH. 


47 


SECTION  IY. — Transverse  Strength — Double  T  and  Hollow 
Rectangle  or  Box  Section. 

44.  Moment  of  Resistance.  Let  Fig.  16  represent 
the  sections  of  the  Double  T  and  Box  Sections  respectively  ;  the 
same  letters  in  the  text  apply  to  both  sections. 


A 


Let  ABDX  represent  the  sections,  OnEMnNO  a  section 
through  the  strain  wedge  on  the  line  OE,  A  OX  the  axes,  and 
nl  the  neutral  lines. 

Notation  for  the  Double  T  : 

~b  =  the  width  of  the  web  in  inches, 
Jj  =  Z>2  =  the  sum  of  the  widths  of  the  flanges  of  the  double 

T  in  inches,  or  AX  —  2>, 

dl  =  the  depth  of  the  top  flanges  of  the  double  T, 
d,  =  the      "       "     "    bottom  «      "    "         "   t     " 

Notation  for  the  Box  Section  : 

1)  —  the  sum  of  the  widths  of  the  sides  of  the  box  in 

inches, 

Jj  =  J2  =  the  inside  width  of  the  box, 
^  =  the  depth  of  the  top  of  the  box, 
d,  =  the      "      "      "   bottonl  of  the  box. 

Giving  the  letters  the  above  definition  the  moments  of  tensile 


48        STRENGTH  OF  BEAMS  AND  COLUMNS. 

and  compressive  resistance  may  be  obtained  from  the  following 
formulas : 

7?c  from  Eq.  34,  (4Y) 

R,     "       "     35.  (48) 

45.  Neutral  Line,     The  position  of  the  neutral  line  in 
the  Double  T  and  Box  Sections,  when  constructed  of  cast-iron, 
may  be  computed  approximately  from  the  formula  given  for 
the  neutral  line  in  the  Hodgkinson  beam,  or 

dQ  from  Eq.  36.  (49) 

"When  beams  of  these  sections  are  made  of  wrought-iron  and 
steel,  the  neutral  line  is  so  close  to  the  bottom  of  the  section 
that  its  position  cannot  be  determined  from  an  approximate 
formula.  An  exact  solution  of  a  long  equation  of  the  fourth 
degree  must  be  made  to  determine  its  position. 

46.  Transverse  Strength,     The   breaking   strength, 
Z,  may  be   computed  from  the  formula  given  for  the  trans- 
verse strength  of  the  Hodgkinson  beam,  by  noting  the  defini- 
tion of  the  letters  given  in  Arts.  35  and  44. 

Z  from  Eq.  37,  (50) 

Z     «       "     38.  (51) 

47.  To  Design  a  Double  T  and  Box  Section, 
Step  I.  Assume  the  depth,  d,  of  the  Box  or  Double  T  and 

the  position  of  the  neutral  line,  also  the  width,  I,  of  the  web 
of  the  Double  T,  or  the  sum  of  the  equal  widths  of  the  sides  of 
the  box.  The  compressed  area  above  the  assumed  position  of 
the  neutral  line  must  sustain  one  half  and  the  extended  area 
below  it  the  other  half  of  the  Bending  Moment  of  the  applied 
load,  Z,  that  the  beam  is  required  to  break  with. 

FOR    THE    COMPRESSED    AREA  \ 

Step  II. — The  moment  of  compressive  resistance  that  the 
sides  of  the  Box  or  the  web  of  the  Double  T  will  offer  to  the 


TRANSVERSE   STRENGTH.  49 

Bending  Moment  of  the  applied  load  from  Eq.  11,  page  10, 
being  substituted  for  R  in  Eq.  22,  page  36,  will  give  for  its 
proportion  of  the  load 

_  mbd^C 

A-     ~67"' 

• 

Step  III.  —  For  the  top  flanges  of  the  Double  T  or  the 
top  of  the  Box  Section  deduct  the  load,  Z15  found  by  Step  II. 
from  one  half  of  the  total  applied  load,  L  ;  the  remainder 
will  be  the  breaking  load,  Z2,  for  the  top  flanges  or  the  top 
of  the  box,  as  the  case  may  require,  or 

A  =  \  -  Lv  (53) 

To  design  an  area  of  section  that  will  resist  this  load,  .Z2, 
assume  a  width,  b19  for  the  sum  of  the  widths  of  the  top  flanges 
or  the  top  of  the  Box,  and  compute  therefrom  their  depth,  dl9 
by  substituting  for  the  moment  of  compressive  resistance,  R, 
in  Eq.  22,  page  36,  its  value  in  this  case  from  Eq.  12,  page  10, 
and  we  will  have 

~  (54) 


from  which  deducing  the  value  of  d1  by  making  d?  =  d*, 
which  may  be  done  without  appreciably  affecting  the  result 
obtained,  we  have 


d        j  e  ( 

9 


from  which  the  required  depth  may  be  computed. 

FOE     THE    EXTENDED    AREA  I 

Step  IV.  —  The  moment  of  the  tensile  resistance  that  the 
web  of  the  Double  T  or  the  sides  of  the  Box  Section  will 
offer  to  the  breaking  moment  of  the  applied  load,  Z,  will  be 


50  STRENGTH   OF   BEAMS   AND    COLUMNS. 

obtained  from  Eq.  13,  page  11,  which  being  placed  for  R  in 
Eq.  22,  page  36,  gives  for  its  proportion  of  the  load,  Z3, 

vn  T 

L,=  MT  (2d  +  dc)~.  (56) 


Step  V.—  The  bottom  flanges  of  the  Double  T  or  the 
bottom  of  the  Box  Section  must  sustain  as  its  proportion  of  the 
load,  Z4, 

L*=\-  A-  (67) 

In  order  to  design  an  area  that  will  sustain  this  load,  Z4, 
assume  a  width,  #a  =  5,,  the  width  assumed  in  Step  III.,  and 
compute  the  corresponding  depth,  d»  by  placing  for  7?,  in 
Eq.  22,  the  value  of  the  moment  of  the  tensile  resistance  for 
this  case,  from  Eq.  14,  page  12,  and  we  have 

-  d.)  -  d:  (3d  -  24)  (68) 


from  which,  deducing  the  value  of  da  by  making  d*  — 
we  have 


,  _         rm-  ImT-ts  m (3c?T -f  3d  +  2)*       /KQN 

262mr(3e^  +  3d  +  2) 

from  which  the  required  depth  may  be  computed. 

Step  VI. — Should  the  section  obtained  by  this  process 
not  be  satisfactory  new  dimensions  must  be  assumed  to  remedy 
the  defects  of  form,  and  the  process  repeated. 

CAST-!RON  DOUBLE  T  AND  Box  SECTIONS  : 

For  the  Box  or  Hollow  rectangular  beam  the  assumed 
widths,  Jj  =  J9,  must  be  placed  between  the  sides  of  the  box, 
that  having  the  depth,  d^  from  Eq.  55,  must  be  placed  at  the 
top,  and  da,  computed  from  Eq.  59,  at  the  bottom  of  the  Box 
beam. 


TRANSVERSE   STRENGTH.  51 

For  the  Double  T  beam  the  assumed  width,  bl  =  &2,  must 
be  placed  in  equal  projecting  flanges  on  each  side  of  the  web 
at  the  top  and  bottom,  that  having  the  depth  d1  at  the  top 
and  dz  at  the  bottom  of  the  beam. 

WROLTGHT-!RON  DOUBLE  T  AND  Box  SECTION. 

For  the  Box  or  Hollow  rectangular  beam  made  of  riveted 
plates  a  part  of  the  assumed  width,  11  =  £>2,  must  be  placed  be- 
tween the  sides  of  the  Box  and  the  balance  in  two  equal  project- 
ing flanges  011  the  outside  of  the  Box  at  the  top  and  bottom  ;  the 
total  width  of  the  rolled  plate  that  forms  the  top  and  bottom 
of  the  Box  is  equal  to  (b  -j-  5,),  and  the  depth  of  the  plates 
that  form  its  sides  is  equal  to  d  —  (dl  -\-  d^)  as  found  by  Steps 
III.  and  Y. 

For  the  Double  T : 

Riveted  plate  sections.  The  width  of  the  top  and  bottom 
plates  is  equal  to  (b  +  &a),  and  the  depth  of  the  web  plate  is 

d-ti  +  d,). 

tolled  Eyebeams  are  arranged  like  that  for  a  cast-iron 
Double  T  beam. 


STRENGTH   OF   BEAMS   AND    COLUMNS. 


SECTION  Y. — Transverse  Strength. 

The  Inverted  T,  Double  Inverted  T  and  [_J  Sections. 

48.  Moment  of  Resistance.  Let  Figs.  17,  18  and 
19  respectively  represent  the  three  sections,  nl  the  neutral 
lines,  and  .4^  the  axes  or  origin  of  moments  of  resistance  for 
each  section;, 


Fxp  17 


A 


-.1 


f 


V 


In  addition  to  the  notation  given   in  Art.   35,  observe  the 
following  for  use  in  the  formulas  given  in  this  Section : 

The  Inverted  T,  Fig.  17. 

b   =  the  width  of  the  web  in  inches, 

Ja  =  the  sum  of  the  widths  of  the  flanges,  or  BD  —  b, 

d^  =  the  depth  of  the  flanges  in  inches. 

The  Double  Inverted  T,  Fig.  18. 

J    —  the  sum  of  the  widths  of  the  webs  in  inches, 

£2  =     "      "        "  "  "      flanges,  or  MN  -  h, 

d^  •=.  the  depth  of  the  flanges  in  inches. 

The  y  Section,  Fig.  19. 

1   =  the  sum  of  the  widths  of  the  webs  in  inches, 

£>2  =  the  width  of  the  bottom  in  inches,  or  OP  —  Z>, 

d2  =  the  depth  of  the  bottom  in  inches. 

Noting  the  above  definitions,  the  moment  of  compressive 
resistance  for  each  section  will  be  from  Eq.  11,  'page  10, 


(60) 


TRANSVERSE   STRENGTH.  53 

and  for  the  moment  of  tensile  resistance  the  formula  given 
for  the  Hodgkinsoii  beam  will  apply,  or 

J??T  from  Eq.  35,  page  42. 
The  Moment  of  Resistance  of  the  section  will  b 

£=fic  +  2tv=  27?c 
from  the  above  equations. 

49.    The  Neutral  Line.     In  cast-iron  beams  of  tin 
sections  the  neutral  line  will  be  given  approximately  by  the 
following  formula  : 

,  _  di  +  vmdf>,d,  (3  +  1)  +  VcH&i+l) 


5O.  Transverse   Strength.      Place  for  It  in  Eq.  22 

the  values  for  the  Moment  of  Resistance  of  the  sections,  and 
we  will  have 


L  =  ,  (64) 

08 

and 

L  from  Eq.  38,  page  44.  (65) 

From  either  of  these  formulas  the  transverse  breaking  load 
may  be  computed. 

51.  To  Design  an  Inverted  T,  Double  In- 
verted T  and  |_|  Sections. 

Step  I.  —  Assume  a  value  for  the  depth,  d,  and  the  width 
or  sum  of  the  widths  of  the  webs,  5,  and  an  economical  posi- 
tion for  the  neutral  line,  and  from  these  compute  the  other 
dimensions. 

Step  II.  —  The  compressed  area  above  the  neutral  line 
must  have  a  moment  of  resistance  equal  to  one  half  the  Bend- 
ing Moment  of  the  applied  load,  Z,  or 

M!C       Zs  ( 

6       "  2m 


54        STRENGTH  OF  BEAMS  AND  COLUMNS. 

If  our  assumed  values  for  b  and  dc  do  not  satisfy  this  equa- 
tion, other  values  must  be  assumed  until  the  two  members 
give  identical  numerical  quantities. 

FOR    THE    EXTENDED    AREA  I 

Step  III.  —  The  load,  Z,  sustained  by  the  extended  area  of 
the  web  or  webs,  #,  may  be  obtained  by  substituting  their  mo- 
ments si  resistance  for  R  in  Eq.  '22,  page  36,  its  value  from 
Eq.  13,  page  11, 

(67) 


Step  IV,  —  For  the  flanges  and  bottom  deduct  the  load  Z: 
found  by  Step  III.,  from  one  half  of  the  applied  load  Z,  the 
remainder,  Za,  is  the  load  that  must  be  sustained  by  the  flanges 
and  bottom. 

Z.  =  |-Z...  .        (68) 

In  order  to  design  an  area  that  will  sustain  this  portion  Z2 
of  the  applied  Z,  assume  a  convenient  depth,  d^  and  from  this 
compute  the  width  62,  by  substituting  for  7?  in  Eq.  22,  page 
36,  its  value,  the  moment  of  tensile  resistance  for  this  case 
from  Eq.  14,  page  12, 


Z2  =     3d,dr  (2d  -  d>)  -  d:  (3d  -  2^)1       —,     (69) 
from  which 


___  ___ 

3d  A  (2d  -  rfj  -  d,2  (3d  -  2d2)  mT  ' 

which  gives  the  required  width. 


TRANSVERSE   STRENGTH. 


SECTION  YI. — Transverse  Strength — Circular  Sections. 

52.  Moment  of  Resistance.  In  Eig.  20,  let  BnOl 
represent  the  section,  BnOMnNB  a  section  through  the 
strain  wedges  on  the  line  BO,  B  the  origin  of  co-ordinates, 
DBS  the  axis  or  origin  of  moments,  nl  the  neutral  line,  nBl 
the  compressed  area,  and  nOl  the  extended  area 


Adopting    the  notation   heretofore   used   and  defined,  we 
have 

C  =  the  greatest  intensity  of  the  compressive  strain  in  pounds, 

pe"r  square  inch, 
T  =  the   greatest   intensity  of  the   tensile  strain  in  pounds, 

per  square  inch, 

J?T  =  the  moment  of  the  tensile  resistance  in  inch-pounds, 
7?c  =  the    moment    of   the    compressive    resistance    in   inch- 
pounds, 

R  =  the  Moment  of  Resistance  of  the  section, 
dc  =  the  versine  Bn  of  one  half  of  the  arc  nBl  in  inches, 
</T  =  the    versine    On    of    one    half    of    the    arc    nOl    in 

inches, 

d  =.  dc  -f-  dT  —  the  diameter  of  the  circle  in  inches, 
r  =  the  radius  of  the  circle  in  inches, 
q  —  the  quotient  arising  from  dividing  C  by  .T7, 


56        STRENGTH  OF  BEAMS  AND  COLUMNS. 

L  —  the  total  applied  load  in  pounds, 
s  =  the  span,  the  distance  between  the  supports 

in  inches, 

m  =  the  factor  denned  in  Art.  34,  page  35, 
Comp.  arc  =  the   arc   bounding    the   compressed   area   in 

inches, 
Tension  arc  —  the  arc  bounding  the  extended  area  in  inches. 

For  the   moment  of    compressive  resistance  from  Eq.  15, 
page  14,  we  have 

c 


+  Comp.  arc  (12dc  -  15r)/],    (71) 
and  for  the  moment  of  tensile  resistance 

*  (6r  -  4)  +  18r'  (r  -  dfi 

+  Tension  arc  (12^T  —  9r)  r9  J,    (72) 


and  for  the  Moment  of  Kesistance  of  the  circular  section, 

27?T.  (73) 


53.  The  Neutral  Line,      The  position  of  the  neutral 
line  in  circular  wooden,  cast-iron,  wrought-iron,  and  steel  beams 
will  be  found  tabulated  on  pages  80,  100  and  112. 

54,  Transverse  Strength,      Substituting  for  R,  in 
Eq.  .22,  page  36,  its  value  2#c  =  27?T  from  Eqs.  71  and  72, 
we  have 


+  Comp.  arc  (12<7C  -  15r)r5,   (74) 


TRANSVERSE   STRENGTH.  57 

and 

mT    ,      .  r^r-  r .  ^  (5/.  _  ^  +  lSr,  (r  _ 


+  Tension  arc  (l2dT  -  9r)  r*    ,    (75) 

from  which  the  breaking  load  of  a  circular  beam  may  be  com- 
puted. By  observing  the  following  equalities,  much  time  and 
labor  will  be  economized  in  comparing  the  results  computed 
from  these  formulas : 


(dc  -  r)  +  r*  (30/-  -  14dc)] 

ISr*  (r  - 


=  ± 
Comp.  arc  =  %7rr  —  Tension  arc. 

Another  Method,  The  Moments  of  Resistance  of 
circular  sections  are  to  each  other  as  the  cubes  of  their  radii  / 
hence  by  computing  and  tabulating  the  Moments  of  Resist- 
ance for  all  required  positions  of  the  neutral  line,  in  a  circle 
whose  radius  is  unity,  those  for  any  other  circle  composed  of 
the  same  material  may  be  computed  by  multiplying  the 
tabular  number  by  the  cube  of  the  radius. 

Let  yc  =  the  second  member  of  Eq.  74,  when  r  =  1,  except 

,,      .,          mC 
the  iactor  ---  , 


,,     ,          mT 
the  factor  --  . 
s 


/T  —  the  second  member  of  Eq.  75,  when  r  =  1,  except 

(76) 
and 

L  =  ?&£.  (77) 


58        STRENGTH  OF  BEAMS  AND  COLUMNS. 

The  application  of  Eq.  76  is  illustrated  in  Examples  11,  32 
and  33,  and  Eq.  77  in  Examples  12,  13  and  14. 

Tables  giving  the  computed  values  of  fc  and  f^  for  the 
different  positions  of  the  neutral  line  in  cast-iron,  wrought- 
iron  and  steel,  circular  sections,  whose  radius  is  unity,  are 
given  on  pages  80  and  100. 

55.  To  Compute  the  Compress!  ve  Strain. 

PROBLEM.  —  The  position  of  the  neutral  line  and  the  com- 
pressive  strain  of  the  material  of  a  circular  section  may  he 
computed  from  the  known  transverse  breaking  load  and  tensile 
strength  of  the  material. 

Deducing  the  value  of  fT  from  Eq.  77,  we  have 


from  the  proper  table  in  the  sequel  take  the  value  of  q  cor- 
responding to  this  value  of  fT  ;  then 

C=qT.  (79) 

Illustrated  by  Examples  3  and  23. 

56,   To  Compute  the  Tensile  Strain. 

PROBLEM.  —  The  tensile  strain  of  the  material  of  a  circular 
section  may  be  computed  from  the  known  transverse  break- 
ing load  and  compressive  strength  of  the  material. 

From  Eq.  76  deduce  the  value  of  f^ 

(80) 

From  the  Table  given  in  the  sequel  for  the  material  take  the 
value  of  £,  corresponding  to  the  computed  value  of  /c,  then 

T=~.  (81) 

2 
Illustrated  by  Example  24. 


TRANSVERSE   STRENGTH.  59 

57.   Relative  Strength  of  Circular  and  Square 

Beams.  Constant  ratios  exist  between  the  Moments  of  Re- 
sistance of  the  circle  and  its  inscribed,  circumscribed  and  equal 
area  square,  when  the  material  composing  the  square  and 
the  circle  has  the  same  tensile  and  compressive  strength,  and 
consequently  the  same  ratios  exist  between  the  transverse 
strength  of  the  beams  of  which  they  are  sections,  the  span  and 
manner  of  loading  the  circular  and  square  beam  being  the 
same. 

General  Formula  : 
Let  f  —  the  position  of  the  neutral  line  in  the  square  when 

the   side  is  unity,  and  d*  —  f*d?,  b  =  d  and  r3  =  —  , 

8 

hence, 

Strength-  of  the  Square  =  ^d'C  =^!^  from  Eq.  23. 


Strength  of  the  Circle  =  r*feC  =       c,  from  Eq.  76, 

8 

from  which  we  have 

Square  :  Circle  :  : 


3  8 

.  •  .  Circle  =  Square  X  —  £|  .  (82) 

3* 


CASE  I.  —  When  the  circle  is  inscribed  within  the  square, 
d  =  d,  and  Eq.  82  becomes 

Circle  =  Square  X  ^4  (83) 

J 

Illustrated  by  Examples  15  and  16  of  the  sequel. 

CASE  II.  —  When  the  square  is  inscribed  within  the  circle  side 
of  the  square,  d  =  diameter  of  the  circle  d  X  0.707,  and  Eq. 
82  becomes 

Circle  =  Square  X  -  (84) 


60 


STRENGTH   OF   BEAMS   AND    COLUIiNS. 


CASE  III.  —  When  the  square  and  circle  are  equal  in  area 
side  of  the  square,  d  =  diameter  of  the  circle  d  X  0.886,  and 
Eq.  82  becomes 

S/. 


Circle  =  /Square  X 


5.564  /' 


Illustrated  by  Examples  18  and  19. 


(85) 


SECTION  VII. — Transverse  Strength— Hollow  Circular 

Sections. 


58.  Moment  of  Resistance,  In  Fig.  21,  let  BnOl 
represent  the  section,  BnOMnNB  a  section  through  the 
strain  wedges  on  the  line  BO,  B  the  origin  of  co-ordinates, 
DBS  the  axis  or  origin  of  moments,  and  nl  the  neutral  line. 


D  S  * 


M 


A 

f 

r 

"t 

T 

c 

*  •*                                                  x 

1 

:_sv.'. 

J- 

Let  ^  =  the  thickness  of  the  metal  ring  in  inches, 
r  =  the  radius  of  the  outer  circumference, 

"     "    inner  « 

T=  T. 


M    /*»  /   it 

ri—  r  —  i  — 


For  other  notation  refer  to  Art.  52. 

When  £,  the  thickness  of  the  metal  ring,  is  very  small,  the 
entire  strain  distributed  over  the  metal  of  the  section  may  be 


TRANSVERSE   STRENGTH.  61 

treated  as  if  it  was  all  concentrated  in  the  outer  surface  of  the 
cylindrical  beam ;  if  t  is  not  very  thin,  r  —  ^  or  the  mean  ra- 
dius must  he  used  instead  of  r  in  the  following  formulas. 

Making  the  substitution  of  tC  for  C  and  tT  for  T  in  Eqs. 
18  and  19,  the  moments  of  compressive  and  tensile  resistance 
become 


^c  =    <T7     Vd*d*  X  2  (3r  -  dc)  +  Comp.  arc  (2dc  -  3r)    ,  (86) 
4acL.  J 

and 

^T  ~  icTL  ^^  X  2  (/>  +  dT)  +  Tension  arc  (2cZT—  r)T(87) 

For  the  Moment  of  Resistance  of  the  section, 

R  =  Rc  4-  J?T  =  27?c  =  2^T.  (88) 

59,  The  Neutral  Line,     The  position  of  the  neutral 
line  in  hollow  circular  sections  of  cast-iron,  wrought-iron  and 
steel,  when  the  radius  of  the  outer  circumference  is  unity,  will 
be  found  tabulated  in  the  sequel. 

60,  Transverse    Strength.      Substituting    for    the 
Moment  of  Resistance  R,  in  Eq.  22,  page  36,  its  values  in  this 
case,  2/?c  =  27?T,  from  Eqs.  86  and  87,  we  have 

mrtCr"     ~| 

Z  =     s£   [    VdcdT  X  2  (3r  —  dc) -\-Comp.  arc  (2dc  —  3r) J?  (o9) 

and 

JL  ^= 


~~\ 

^  X  2  (/>  +  ^T)  +  Tension  arc  (2dT-  r) J.  (90) 

From  either  of  the  above  formulas  the  transverse  strength  of 
hollow  cylindrical  beams  may  be  computed. 


62        STRENGTH  OF  BEAMS  AND  COLUMNS. 

In  these  formulas  the  following  equalities  exist  : 

2  (3r-  dc)  =   Vd^dT  X  2  (/•  +  d, 


Comp.  arc  =  'Znr  —  Tension  arc. 

When  the  metal  ring  is  very  thin,  r  in  the  above  formulas 
is  the  radius  of  .the  outer  circle,  otherwise  it  is  the  radius  of 

the  mean  circle,  r  —  —  . 

Another  Method,  The  Moments  of  Resistance  of  thin 
hollow  circular  sections  are  to  each  other  as  the  squares  of 
their  radii  ;  hence,  by  computing  and  tabulating  the  Moments 
of  Resistance  for  all  required  positions  of  the  neutral  line,  in  a 
thin  hollow  circle  whose  radius  is  unity,  those  for  any  other 
thin  hollow  circle  composed  of  the  same  material  may  be 
computed  by  multiplying  the  tabular  number  by  the  square  of 
the  radius. 

e  —  the  second  member  of  Eq.  89,  when  radius  —  1,  ex- 

u  x-i     £          mtC 
cept  the  factor  --  , 


fT  —  the  second  member  of  Eq.  90,  when  radius  =  1,  ex- 

.  ,,     ,          mtT 
cept  the  tactor  -  . 

s 

With  this  notation  Eqs.  89  and  90  become 


and 

L  =  ?.  (92) 


TRANSVERSE   STRENGTH.  63 

Tables  giving  the  computed  values  of  fc  and  fT  for  thp  posi- 
tions of  the  neutral  line  in  thin  hollow  circles,  whose  radius  is 
unity,  will  be  found  in  the  sequel. 

The  application  of  the  above  formulas  is  illustrated  in 
Examples  20  and  34 


CHAPTEK  IY. 

CAST-IRON   BEAMS. 
SECTION  I. —  General  Conditions. 

61.  Compressive  Strength.     Crushing. — The  crush- 
ing strength  of  cast-iron  that  is  usually  obtained  by  experi- 
menters and  recorded  for  use  in  designing  structures,  is  the 
number  of  pounds   avoirdupois   that   it   requires   to  crush  a 
prism  of  the  material  whose  sectional  area  is  one  square  inch 
and  length  from  one  and  a  half  to  three  times  the  diameter, 
under  which  condition  it  is  found  to  be  more  nearly  constant 
in  value  for  the  same  material  than  when  the  height  bears  a 
greater  or  less  ratio  to  the  least  diameter  of  the  prism  tested. 

Value. — The  range  of  values  for  the  crushing  strength  of 
cast-iron  may  be  taken  as  being  from  85000  pounds  to  125000 
pounds  per  square  inch  ;  the  mean  is  about  100000  pounds. 

Elastic  Limit. — The  compressive  elasticity  of  cast-iron  as 
recorded  by  the  earlier  experimenters  appears  to  be  very 
defective,  but  improved  methods  of  manufacture  have  pro- 
duced a  cast-iron  from  which  modern  experimenters  find  the 
increase  in  the  amount  of  the  compression  to  be,  practically, 
in  direct  proportion  to  the  increase  in  the  load,  within  the 
elastic  limit  of  the  cast-iron. 

The  compressive  elastic  limit  varies  from  three  fifths  to 
nearly  the  crushing  strength. 

62.  Tensile  Strength.     Tenacity  is  the  force  in  pounds 
that  is  required  to  pull  asunder  a  prism  of  cast-iron  whose  sec- 
tional area  is  one  square  inch.     It  ranges  in  value  from  15000 
to  30000  pounds. 


GENERAL   CONDITIONS.  6fr 

Elastic  Limit. — The  extensions  within  the  elastic  limit  fol- 
low laws  similar  to  those  that  govern  the  compressions.  The^ 
elastic  limit  is  about  one  half  of  the  tensile  strength. 

63,  Ratio  of  the  Compressive  to  the   Tensile 
Strength.     Experiments  have  demonstrated  that  the  ratio 
existing  between  the  crushing  and  tensile  strength  of  cast-iron 
has  a  wider  range  of  values  than  that  for  any  other  known 
material,  and  that  this  results  from  great  fluctuations  in  the 
crushing  rather  than  in   the   tensile  strength.     The  extreme 
values  for  the    C  -f-  T  =  q  of  our  formulas  may  be  taken  at 
from  3  to  8J ;  the  tendency  of  improved  methods  of  manu- 
facture is  to  decrease,   numerically,   this  ratio   by  increasing 
that  of  T  and  decreasing  that  of  C\  which  in  pure  orwrought- 
iron  becomes  C  -=-  T  =  1. 

64.  Transverse  Strength.      Cast-iron  breaking  with 
a  well-defined  fracture,  its  transverse  strength  may  be  com- 
puted   from   our   formulas   when   the    crushing   and    tensile 
strength   of   the   identical   cast-iron   composing   the  beam  is 
known  from  experiment,  as  the  values  of  C  and  T  are  found 
to  vary  in  an   uncertain  manner,   with  remelting,  length   of 
time  in  fusion,  etc. 

The  transverse  strength  of  some  cast-iron  does  not  increase 
directly  with  the  increase  in  the  dimensions  of  its  cross-section, 
as  it  should  in  accordance  with  well-defined  laws,  but  in  a 
lower  ratio.  This  defect  is  usually  imputed  to  unequal  strains 
being  brought  upon  the  metal  in  different  parts  of  the  section, 
from  unequal  temperatures  in  cooling,  but  few  experiments 
have  been  made  to  test  the  matter. 

Col.  James,  of  England,  planed  out  and  tested  a  f-inch 
square  bar  from  a  2-inch  square  cast  bar,  the  values  of  C  and 
Tior  this  brand  of  iron,  Clyde  No.  3,  having  been  determined 
from  experiments  by  Mr.  Eaton  Hodgkinson.  The  following 
is  the  result  of  Col.  James's  tests : 


66        STRENGTH  OF  BEAMS  AND  COLUMNS 

Clyde  No.  3.  The  Dressed  Bar.  Transverse  Load. 

O=  106039  Ibs.         C  =  60233  Ibs.  by  test.  193  Ibs.  rough  bar. 

T=  23468    "          T  —  14509  "     by  computation.*       134    "     planed  out. 

From  the  above  it  appears  that  the  value  of  C  decreased  in 
the  "  planed-out"  bar  56.8  per  cent ;  that  of  T  60.5  per  cent, 
and  the  transverse  breaking  load  69.4  per  cent.  But  this 
result  is  contrary  to  that  obtained  in  the  following  tests  made 
for  the  United  States  Government — Report  of  Tests  of  Iron  arid 
Steel  for  1885. f  In  the  United  States  Government  tests  the 
bars  were  cast  two  feet  in  length  and  three  inches  in  width, 
with  depths  ranging  from  a  half  to  two  inches.  The  equiva- 
lent centre  breaking  load  for  a  bar  of  the  metal  one  inch 
square  and  twelve  inches  span  was  computed  from  the  result 
of  each  experiment,  that  they  might  be  compared. 

Condition.    Rough.  Edges  Planed.  Edges  Planed.       Planed  all  over. 

No.  Expts.        4  3  6                               6 

Size,           3"  X  0".5  2"  X  1"  to  1".28  2"  X  1".5  to  2"    2"  X  1".S5  to  1".75 

Load,  Ibs.,      2453  2136  2025                          2134 

The  "  edges  planed"  bars  were  reduced  in  width  from  the 
rough  castings  that  were  3"  wide,  the  tension  and  compression 
surfaces  being  left  as  they  were  cast. 

The  "  planed-all-over"  bars  were  cut  from  cast  bars  that 
were  three  inches  wide  and  two  inches  deep,  equal  depths  of 
metal  having  been  removed  from  each  face  of  the  rough  bar. 

In  these  experiments  the  strength  of  the  bars  that  were  cast 
two  inches  deep  varied  about  five  per  cent  from  those  cast 
one  inch  deep,  and  the  "  planed-out"  bars  were  as  strong  as 
the  rough  cast  bars  of  the  same  size. 

The  bars  that  were  tested  with  the  tension  and  compres- 
sion surfaces  as  cast  varied  in  strength  about  5  per  cent  from 
the  average  strength  given  above,  and  the  "  planed-out "  bars 

*  Example  No.  4. 

f  Senate  Ex.  Doc.  No.  36— 49th  Congress,  1st  Session,  p.  1162. 


GENERAL   CONDITIONS.  67 

only  2J-  per  cent  in  strength  from  the  averages  given,  indicat- 
ing great  uniformity  in  strength,  and  consequently  no  de- 
crease in  the  values  of  C  and  T  for  the  metal  in  the  interior 
of  the  cast  bar,  but  this  was  not  determined  by  direct  exper- 
iment. 

The  following  ratios  are  usually  quoted  to  illustrate  the  de- 
preciation of  the  transverse  strength  of  cast-iron  beams  re- 
sulting from  an  increase  in  the  size  of  the  cross-section,  the 
strength  that  the  large  beam  should  have  being  computed 
from  that  of  the  V  X  1"  bars,  the  latter  being  denoted  by 
100. 

Experimenter.  1"  X  1"        2"  X  2"          3"  X  3" 

Capt.  James 100  71.84          61.95  per  cent, 

Mr.  Hodgkinson..          100        -    71.22  " 

The  transverse  elastic  limit  load  may  be  computed  when 
the  corresponding  values  of  C  and  T  have  been  determined 
by  experiment. 

65.  To  Compute  the  Compressive  and  Tensile 
Strength.  The  relation  existing  between  the  Transverse 
Load,  Compressive  and  Tensile  Strength  is  such  that  the 
value  of  any  one  of  the  three  may  be  computed  when  the 
value  of  the  other  two  has  been  determined  by  direct  experi- 
ment ;  this  relation  is  true  for  both  the  elastic  limit  and  the 
breaking  values. 

The  beam  from  which  the  experimental  breaking  load  was 
obtained  must  have  been  sufficiently  long  to  deflect  a  distance 
equal  to  or  greater  than  the  depth  of  tension  area  that  is  re- 
quired for  true  cross-breaking,  or  else  our  computed  value  of 
C  or  T  will  be  incorrect  for  the  reason  given  in  Art.  24. 

Compressive  Strength.—  This  may  be  computed  from  the 
known  tensile  strengtii  of  the  material  and  the  transverse 
breaking  load  of  rectangular  and  circular  beams,  or  of  that 
of  any  of  the  flanged  beams  when  the  neutral  line  lies  within 
the  lower  or  tension  flange. 


68        STRENGTH  OF  BEAMS  AND  COLUMNS. 

The  cast-iron  that  Mr.  Barlow  used  in  his  experiments  was 
composed  of  pig  and  scrap-iron  ;  he  determined  its  tensile  and 
transverse  strength,  but  he  does  not  record  in  his  "  Strength 
of  Materials  "  its  crushing  strength  ;  this  we  can  compute  from 
the  data  given. 

EXAMPLE  1.  —  Required  the  Crushing  Strength  of  Mr.  Bar- 
low's iron  from  the  centre-breaking  transverse  load  of  a  rect- 
angular beam  when 

The  depth  .  .d  =  1  inch,       T  —  18750  pounds  by  experiment, 
"    breadth  I  =  1.02  inch,  L  =      534        "  "        * 

"    span  .  .  .  s  =60.  inches,  m  —         4. 

The  position  of  the  neutral  line  must  be  computed  from 
Eq.  30,  page  40, 

,  _  3  X  1  /9  X  4  X  1.02(1)*  i«750  -  12  X  584  X  60       n,, 

2        "  V  '  4  X  4  X  1.02  X  18750  °  - 

hence  dc  =  1  -  0.5032  =  0".4968, 

from  which  we  can  compute  the  value  of  the  crushing  strain 
by  means  of  Eq.  31, 


The  following  Example  is  taken  from  Major  Wade's  Ex- 
periments on  the  "  Strength  and  other  Properties  of  Metals 
for  Cannon,"  made  for  the  United  States  Government  : 

EXAMPLE  2.  —  Required  the  Crushing  Strength  of  Major 
Wade's  cast-iron  in  third  fusion,  from  the  transverse  strength 
of  a  rectangular  beam,  when 

The  depth.  ..d  =  2.01    inches,  T  =  26569  pounds  by  test, 
"    breadth  I  —  2.008       "       L  =  16172       "        "      " 
"    span.  .  ..«  =  20.  "        m  =  4          for  a  centre  load. 

*  Barlow's  "  Strength  of  Materials,"  p.  152. 


GENERAL   CONDITIONS.  69 

For  the  position  of  the  neutral  line,  we  have  from  Eq.  30, 


3X2.01  _       /9~X  4x  2.008(2.01)^^6^"l2~Xl6172  X  20  _  ft,, 
V  '  4  X  4  X  2.008  X  26569 

.  • .  dc  =  2.01  —  0.8835  =  1.1265  inches. 


Then  we  can  compute  the  crushing  strength,  (7,  from  Eq. 
31, 

n         3  X  16172  X  20 

~-  4  X  2.008  (1.1265?  =  952°°  P°Unds' 

EXAMPLE  3.  —  Kequired  the  Crushing  Strength  of  certain 
cast-iron  from  the  transverse  strength  of  a  circular  beam 
when 

The  diameter  d  —  1//.129,    T  —  29400  pounds  mean  of  tests, 
'•    span  .....  *  =  20".0,       L  =  2118        "        by  test, 

m  =  4. 


From  Eq.  78  we  have 

-  2H8  X  20 


4  (0.5645)3  29400 


~ 


From  the  Table,  page  80,  for  the  above  value  of  fT,  we  have 
q  =  3.275,  and  from  Eq.  79, 

.  • .  O=  29400  X  3.275  =  96285  pounds. 

From  the  United  States  Government  Report  of  the  Tests  of 
Iron  and  Steel  for  1884,*  the  mean  crushing  strength  of 
this  cast-iron  was  C  —  100700  pounds. 

The  Crushing  Strength  may  be  computed  from  the  Trans- 
verse Strength  of  the  Hodgkinson  or  any  other  flanged  beam. 
When  the  neutral  line  is  not  situated  above  the  top  line  of  the 
tension  flange,  its  position  may  be  computed  from  Eq.  30, 
page  40,  as  if  the  beam  were  rectangular.  When  the  neutral 

*  Senate  Ex.  Doc.  No.  35— 49th  Congress,  1st  Session,  p.  284. 


70        STRENGTH  OF  BEAMS  AND  COLUMNS. 

line  is  located  above  the  top  line  of  the  tension  flange,  its 
position  can  be  determined  by  the  method  given  in  Problem  I., 
page  43,  but  in  either  case  the  value  of  C  must  be  computed 
by  formula  37,  page  44,  C  being  then  the  only  unknown 
quantity  that  it  will  contain. 

Tensile  Strength.  —  This  may  be  computed  from  the  known 
Crushing  Strength  of  the  material  and  the  Transverse 
Strength  of  any  rectangular,  circular  or  flanged  beam. 

EXAMPLE  4.  —  Required  the  Tensile  Strength  from  the  Trans- 
verse Strength  of  a  rectangular  beam  made  of  a  certain  kind 
of  cast-iron  that  was  tested  by  Captain  James,  of  England,* 
when 

The  depth     d  =  0.75  inches,   O=  60233  pounds  by  test, 
"   breadth  b  =  0.75       "       L  =  134  "        "       « 

"   span        s  =  54.0       "       m  =  4 

The  position  of  the  neutral  line  is  to  be  computed  from  Eq. 
32,  page  40, 


,  / 

'  ==  * 


3  X  134X54 


4X07X0288         ' 
hence  dT  =  0.75  —  0.4  —  0.35  inches. 

Then  we  compute  the  required  value  of  the  tensile  strength, 
T,  from  Eq.  33,  page  40, 

3  X  134  X  54 
T  :  =  4x0.75x0.85(2x0.75+0.4)  ==  14509 

Mr.  llodgkinsoii  determined  for  this  iron, 

Clyde,  No.  3,  C  =  106039  and  T  =  23468  pounds. 


*  British  "  Report  on  the  Application  of  Iron  to  Railway  Structures," 
p.  257. 


RECTANGULAR   CAST-IRON   BEAMS. 


71 


SECTION  II. — Rectangular   Cast-Iron  Seams. 

66.  The  Neutral  Line.  The  position  of  the  neutral 
line  in  rectangular  cast-iron  beams,  for  the  different  required 
ratios  of  C  -=-  T  =  q,  have  been  computed  from  Eq.  26,  page 
38,  and  tabulated  below  for  reference. 

TABLE  of  positions  of  the  neutral  line  in  rectangular  cast- 
iron  beams. 


Ratio  of  Crushing 
to  Tenacity 
or 

Depth  of  Neutral  Line 
Below  the  Crushed  Side 
of  the  Beam,  or 

Ratio  of  Crushing 
to  Tenacity, 
or 

Depth  of  Neutral  Line 
Below  the  Crushed  Side 
of  the  Beam,  or 

•0-*-  T=q. 

dc. 

C+.T=q. 

(lc. 

8.0 

0.4191  d 

5.5 

0.4831  d 

7.875 

0.4127  d 

5.375 

0.4871  d 

7.75                     0.4243  d 

5.25 

0.4913  d 

7.625                    0.4226  d 

5.125 

0.4956  d 

7.5                        0.4298  d 

5.0 

0.5000  d 

7.375                    0  4326  d 

4.875 

0.5045  d 

7.25                     0.4354  d 

4.75 

0.5092  d 

7.125 

0.4384  d 

4.625 

0  5139  d 

7.0 

0.4414  d 

4.5 

0  5189  d 

6.875 

0.4444  d 

4.375 

0.5240  d 

6.75 

0.4475  d 

4.25 

0.5293  d 

6.625 

0.4507  d 

4.125 

0.5347  d 

6.5 

0.4540  d 

4.0 

0.5403  d 

6.375 

0.4573  d 

3.875 

0.5461  d 

6.25                      0.4608  d 

3.75 

0.5521  d 

6.125                    0.4642  d 

3.625 

0.5583  d 

6.0 

0.4678  d 

3.5 

0  5647  d 

5.875 

0.4715  d 

3.375 

0.5714  d 

5.75 

0.4754  d 

3.25 

0.5784  cZ 

5.625 

0.4791  d 

3.125 

0.5855  d 

3.0 

0.5930  d 

Should  lower  numerical  values  of  q  be  required,  refer  to  the 
table  of  neutral  lines  in  rectangular  Sections  of  Steel. 

67.  Transverse  Strength.  From  either  Eq.  27  or 
28,  as  may  be  most  convenient,  the  transverse  strength  of  rect- 
angular cast-iron  beams  may  be  computed. 


72        STRENGTH  OF  BEAMS  AND  COLUMNS. 

EXAMPLE  5. — Required  the  centre  breaking  load  of  a  rect- 
angular cast-iron  beam  when 

The  depth d  =   1  inch,      C  =  102434  pounds  by  test, 

"     breadth b=    1    "         T=  16724         " 

"     span s  =  54  inches,  q  —  6 . 125  and  m  =  4. 

The   position  of  the  neutral  line  may  be  computed  from 
Eq.  26, 


1(_  o.5  +  1/2  X  6.125+2.25)  . 

*<  =  6.125  +  1 

or  its  value  may  be  taken  from  the  Table.  . 

The  transverse  breaking  load,  Z,  from  Eq.  28,  becomes 


L  =  .  646 

3  X  54 


Mr.    Hodgkinson,   in   his   experiments, 
gives   the  above  values  of    C  and   T  for 
Blaen-Avon  No.  2  iron,  and  447  pounds 
.,  «*     as   the   mean    transverse    strength    of    4 

beams  tested,  while  Captain  James  found 
556  pounds  to  be  the  mean  of  3  beams 


of  the  above  dimensions  for  this  brand  of 
iron. 

EXAMPLE  6. — Required  the  centre  breaking  load  of  the  fol- 
lowing :  1-inch  square  cast-iron  beams,  the  span  being  54 
inches,  and  the  values  of  (7,  77and  the  experimental  break- 
ing loads  being  from  Mr.  Hodgkinson' s  experiments.  It  is 
very  probable  that  these  bars  did  not  deflect  sufficiently  to  de- 
velop the  true  transverse  strength.* 

*  Barlow's  "  Strength  of  Materials,"  p.  163. 


RECTANGULAR   CAST-IRON   BEAMS. 


73 


TESTED  STRENGTH  PER 
SQUARE  INCH. 

Neutral 
Line,  or 

rfc. 

TRANSVERSE 

STRENGTH. 

Crushing  C. 

Tensile  T. 

Computed. 

Tested. 

Carron  No.  2,  C   B.  . 
"     2,  H.  B. 
«'    3,  C.  B.. 
"     3,  H.  B. 
Buffery  "     1,  C.  B.. 
"     1,  H.  B. 

Alccin 

106375 
108540 
115442 
133440 
93366 
86397 

16683 
13505 
14200 
17755 
17466 
13434 

0.457 
0.418 
0.416 
0.427 
0.488 
0.456 

Ibe. 
548 
468 
493. 
606 
549 
429 

Ibs. 
476 
463 
446 
527 
463 
436 

468 

515 

EXAMPLE  7. — Required  the  centre  breaking  load  of  a  rect- 
angular cast-iron  beam,  from  Mr.  Barlow's  experiments, 
when 

The  depth .  .  d  =    2.00  inches,  0  —  95462  Ibs.,  computed,  Ex.  1, 
"     breadth  b=    1.99     "       T=  18750    "    by  test, 
"     span s  =  60.00     "        q  =  5 . 091  and  m  =  4. 

For  the  neutral  line  we  have,  from  Eq.  26,  dc  =  0".9934, 

Computed  transverse  strength  from  Eq.  27  or  28 ...  .4175  Ibs., 
Tested  "      3863    « 

68.  To  Design  a  Rectangular  Cast-Iron  Beam. 

The  principles  and  formulas  required   are  given  in  Art.  38, 
page  39. 


74        STEENGTH  OF  BEAMS  AND  COLUMNS. 


SECTION  III. — HodgWmson  Cast-iron  Beams. 

69.  In  this  section  we  will   consider  the  principles  as   if 
applied    to   that   class   of   cast-iron   beams  whose    bottom  or 
tension   flange   is   larger   than  its   top   or  compressed  flange, 
though  they  are  applicable  to  all  other  forms  of  flanged  beams. 

70.  Mr.  Hodgkinson's  Experiments.      Previous 
to  the  year  1840,  when  Mr.  Hodgkinson  commenced  his  ex- 
perimental investigation  into  the  strength    of   materials,   the 
only  metal  beam  that  had  been  used  in  structures  to  any  ex- 
tent was  the  Inverted  T,  Fig.  17,  and  the  Double  T,  Fig.  16, 
in   which   the   top   and    bottom   flanges   were  equal  in  area. 
Commencing  his  experiments  with  the  equal  flanged  Double 
T,  he  f  ound  that,  by  increasing  the  area  of  the  lower  or  tension 
flange    by    small    amounts,    he    continued    to    increase    the 
transverse    strength   of   the  beam,   his   unit  of   measure  and 
standard  of  comparison  being  the  quotient  obtained  from  di- 
viding the  breaking  load  by  the   area  of  the  section  of  the 
beam  expressed  in  inches,  and  that  this  continued  to  increase 
until  he  had  reached  the  point  where  the  tension  flange  was 
about  six  times  the  area  of  the  compressed  flange ;  increasing 
it  to  a  greater  ratio  he  found  that  the  transverse  strength  per 
square  inch  of  section  began  to  decrease.     Experimenting  with 
cast-iron,  in  which  the  ratio  of  C  to  jT,  or  the  crushing  to  the 
tensile  strength,  was  about  siae,  he  recommended  that  the  ten- 
sion flange  should  have  six  times  the  area  of  the  compressed 
flange,  in  order  to  obtain  the  greatest  transverse  strength  per 
square  inch  of  section.     But  subsequent  iifvestigators,  experi- 
menting with   cast-iron   possessing   greater  tenacity,  or  a  less 
ratio  of  C  to  T  than  that  used  by  Mr,  Hodgkinson,  found  the 
greatest  strength  with  a  lower  ratio  of  extended  to  compressed 
flange  than  that  recommended  by  him.     The  reason  for  this 
will  be  apparent  when  the  formulas  for  the  strength  of  these 
beams  are  examined. 


HODGKINSON    CAST-IRON   BEAMS.  75 

71.  Neutral  Line.     When  the  neutral  line  is  not  situ- 
ated within  the  tension  flanges,  its  position  may  be  approxi- 
mately computed  from  Eq.  36,  page  42.     Should  the  neutral 
line  lie   within  the  tension  flange,  its  position  may  be  accu- 
rately computed  from  Eq.  30,  or  from  the  methods  given  in 
Probs.  1  and  2,  page  43,  when  the  transverse  load  and  either 
the  compressive  or  tensile  strength  has  been  determined  from 
experiments.  ___________ 

'M-ET        ~~"1 

72.  Transverse  Strength.    The  '  _T 

transverse  strength  of  the  Hodgkinson  or  Af 

of  any  beam  having  flanges  at  the  top  .v 

and  the  bottom  may  be  computed  from  . —  \ 

either  Eq.  37  or  38,  as  may  be  most  con-  f^ 
venient. 

EXAMPLE  8. — Required  the  centre  breaking  load  of  a  Hodg- 
kinson cast-iron  beam  when 

The  depth  of  the  beam d=  5.125  ins.,   C  =  96000  Ibs., 

"   breadth  of  the  web I  =  0. 34     «      T  =  16000  " 

"         "  "     top  flanges,  ^  =  1.22      "        q  =  6 

"    depth          "      "         "       d,  =  0.315    "        s  =  54  ins. 

"  bottom  "       d,  =  0.56      "      m  =  4. 
"   breadth       "        "       "       b,  =  4.83      " 

Neutral  Line. — Our  formula  36,  giving  the  position  of  the 
neutral  line  approximately  only,  places  it  in  each  case  a  little 
too  near  the  tension  flange.  In  this  example  it  locates  it 
within  the  tension  flange,  and  we  will  assume  that  it  coin- 
cides with  its  upper  line,  or 

dc  =  4".565  and  dT  =  0.56. 
From  Eq.  37  the  transverse  breaking  load  becomes 

_  0.34  (4.565)3+1.22  (0.315)2 (3x4.565-2X0.315)  _ 

3X4565X54  ~ 

The  tension  area  being  continuous  from  the  neutral  line  to  the 


T 


76         STRENGTH  OF  BEAMS  AND  COLUMNS. 

bottom  of  tlie  beam,  Eq.  28,  page  38,  can  be  used  to  compute 
the  load  from  tlie  tensile  strength  T,  in  which  Z>  =  Z>  +  62  of 
this  example. 

L  =  4X5.17X0.56  (2X5.125+4.565)  16000  _ 
3X54 

These  two  values  of  L,  not  being  identical  in  numerical  value, 
our  assumed  position  for  the  neutral  line  must  have  been  a 
little  too  low  down  in  the  section  for  exact  equilibrium. 

The  above  described  beam  was  one  of  the  series  of  beams 
that  was  used  by  Mr.  Hodgkinson  to  determine  the  "  section 
of  greatest  strength."  Its  transverse  breaking  load  was,  by 
test,  16730  pounds.*  The  value  for  T  that  we  use  was  also 
determined  from  experiment ;  the  ratio  was  supposed  to  be 
O  -r-  T  =  6.  In  this  series  of  beams  the  neutral  line  of 
rupture  was  continuously  lowered  in  the  section,  from  pro- 
gressively increasing  the  area  of  the  tension  flange  until  it 
finally  reached  the  upper  surface  of  this  flange. 

EXAMPLE  9. — Required  the  centre  breaking  load  of  a  Hodg- 
kinson Cast-Iron  Beam  (Fig.  15),  when 

The  depth  of  the  beam  d  =  14.0  ins.,  C  =  75983  Ibs.  by  test, 

"   breadth  of  the  web b=    1.0"      2—13815   "     "     " 

"       top  flanges d\  =     1.0  "       q  =  5.5, 

"depth        "         "         "       b,=    2.5"       «  =  16  feet, 

bottom  flanges d*  —    1.75"     m  =  4. 

"    breadth     "  "  "      ....&„=  11.00 " 

From  Eq.  36  the  position  of  the  neutral  line  becomes 


,        1X14+^12X14X11X1.75  (5.5  +  1)  + (1X14)*  (4*5  5+5) 

'    2X1(5.5+!)-  =  13.5ms., 

which  is  within  the  tension  flanges,  assuming  dc  -—  14  —  1.75  = 
12.25. 


*  Barlow's  "  Strength  of  Materials,"  p.  lr 


HODGKINSON   CAST-IRON   BEAMS.  77 


From  Eq.  37  the  breaking  load,  Z,  becomes 


_          .  .)*  (3X12.25-2X1)  _ 

3X12.25X12X16 

From  the  tensile  strength,  T,  and  Eq.  28,  L  becomes 


L  =  JAl2  ><  1-rcPX  M  +  l&gg  =  80109  Ib, 

From  this  near  identity  of  values  for  Z  we  conclude  that  our 
assumed  position  for  the  neutral  line  was  very  near  the  correct 
position. 

Two  of  these  beams  were  constructed  of  Calder  No.  1  cast- 
iron,  and  broken  with  centre  loads  of  73920  and  76160  pounds 
respectively  by  Mr.  Owens,  Inspector  of  Metals  for  the  British 
Government.*  The  values  of  C  and  T  were  determined  by 
Mr.  Hodgkinson  for  this  brand  of  iron.  The  bottom  table 
contained  six  times  the  area  of  the  top  table,  as  recommended 
by  Mr.  Hodgkinson. 

These  beams  broke  with  the  deflections  l//.87  and  2".0  re- 
spectively, and  the  full  strength  of  the  section  was  thus 
developed  by  the  transverse  load. 

The  following  statement  gives  the  proportion  of  the  load 
sustained  by  each  member  of  the  transverse  section  and  the 
proportion  of  the  area  that  it  occupied  : 

Load.  Area. 

Compressed  Flanges  ............      2.3  7.0  per  cent. 

"  Web  .............   47.7  34.3          " 

Extended          «    .............      4.2  5:0  « 

"  Flanges  ...........   45.8  53.7 


100.0  100.0 

The  web  extends  through  the  total  depth  of  the  beam. 

*  Box's  "  Strength  of  Materials,"  p.  202. 


78  STRENGTH    OF   BEAMS   AND    COLUMNS. 

73.   To    Design    a    Hodgkinsoii    Beam.      The 

formula  required   in  designing  a  Uodgkinson  beam  will  be 
found  in  Art.  42,  page  44. 


SECTION  IV. — Double  T  and  Box  Cast-iron  Beams. 

74.  The   Neutral   Line.     The  position  of  the  neutral 
line  may  be  approximately  computed   from   Eq.  36,  page  42, 
by  giving  the  letters  of  the  formula  their  definitions  in  Art. 
44. 

75.  Transverse   Strength.    The  Transverse  Strength 
,.. , — y..      may  be  computed  from  Eqs.  37  and  38, 

I  I ,     , ^J  I      giving  the  letters  their  definition  in  Art. 

*       i  44' 

.  _.x'  EXAMPLE    10. — Required    the    centre 

|i—  i         breaking  load  of  a  Hollow  Rectangular 
\y\         Cast-Iron  Beam  whose  outside  dimension 


i  I     

is    3".125   X    3".125,    inside    2".375  X 

2".375,  thickness  of  metal  all  around  0.375,  when 

The  depth  of  the  beam d  =  3".125          C  =  84000  Ibs. 

"    breadth     "  sides b  =  2  X  0".375  T—  14000   " 

"          "          "  top .b,  =  27/.375          q  =  6, 

"    depth        "  «      d,  =  0^.375          s  =  6  feet, 

"          "          "  bottom  ..  .d9  =  0".3Y5          m  =  4. 
"    breadth     «  «        ...ft,  =  2".375 

The  position  of  the  neutral  line  dc  from  Eq.  36  becomes 




3.125X0.75+  4/12X3.125  X  0.75  X  2.375  (6+1)  +  (3. 125+0/75)2  (4x6+5)= 
2X0.75(6  +  !) 

1".965. 


DOUBLE   T   AND   BOX   OAST-IKON   BEAMS.  79 

The  breaking  load  from  Eq.  37  becomes 

0.75  (1.965)* +2.375  (0.375)'  (3X1.965-2X0.375)  X4x84000_5760 

3  X  1.965  X  12X6  lbs" 

and  from  the  tensile  strength  T7,  and  Eq.  38,  L  becomes 
L  =  5178  pounds. 

In  the  investigations  preliminary  to  the  construction  of  the 
Menai  Straits  Tubular  Bridge,  Mr.  Stephenson  broke  this 
beam  with  5387  pounds.* 

76.  To  Design  a  Double  T  and  Hollow  Rectangular  Cast- 
iron  Beam.  This  may  be  done  by  using  the  formula  and 
directions  given  in  Art.  47,  page  48. 

*  "  Britannia  and  Con  way  Tubular  Bridges,"  p.  429. 


80 


STRENGTH   OF   BEAMS    AND    COLUMNS. 


SECTION  Y. — Circular  Cast- Iron  Beams. 

7  7 .  Neutral  Line.     Table  of  positions  of  the  neutral  line 
in  Circular  Cast-Iron  Beams  and  factors  for  use  in  Eqs.  76  and  77. 


Ratio  of  Crushing 
to  Tenacity,  or 
C  -t-  T  =  q 

Depth  of  Neutral 
Line  Below  the 
Crushed  Side  of  the 

FACTORS  FOR  COMPUTING  THE  MOMENT 
OF  RESISTANCE,  r  —  1. 

Beam,  or  dc. 

fc  for  Crushing 
Strain,  (7. 

ft  for  Tensile 
'    Strain,  T. 

8. 

0.3984  d 

0.3226 

2.5804 

7.875 

0.4004  d 

0.3260 

2.5722 

7.75 

0.4022  d 

0.3298 

2.5640 

•7.625 

0.4041  d 

0.3336 

2.5556 

7.5 

0.4062  d 

0.3380 

2.5466 

7.375 

0.4087  d 

0.3426 

2.5356 

7.25 

0.4112.  d 

0.3472 

2  5250 

7.125 

0.4137  d 

0.3518 

2  5138 

7.0 

0.4161  d 

0.3568 

2.5030 

6.875 

0.4186  d 

0.3620 

2.4920 

6.75 

0.4201  d 

0.3672 

2.4808 

6.625 

0.4236  d 

0.3726 

2.4694 

6.5 

0.4262  d 

0.3780 

2.4578 

6.375 

0.4289  d 

03836 

2.4458 

6.25 

0.4316  d 

0.3892 

2.4346 

6.125 

0.4344  d 

0.3952 

2.4208 

6.0 

0.4373  d 

0.4014 

2.4080 

-  5.875 

0.4402  d 

0.4078 

2.3952 

5.75 

0.4432  d 

0.4142 

2.3812 

5.625 

0.4463  d 

0.4210 

2.3664 

5.5 

0.4494  d 

0.4278 

2  3536 

5.375 

0.4526  d 

0.4350 

2.3392 

5.25 

0.4559  d 

0.4426 

2.3240 

5.125 

0.4594  d 

0.4504 

2.3080 

5.0 

0.4629  d 

0.4584 

2.2916 

4.875 

0.4665  d 

0.4668 

22750 

4.75 

0.4702  d 

0.4752 

2.2580 

4.625 

0.4740  d 

0.4842 

2.2400 

4.5 

0.4780  d 

0.4938 

2.2218 

4.375 

0.4821  d 

0.5036 

2.2028 

4.25 

0.4863  d 

0.5138 

2.  1832 

4.125 

0.4906  d                      0.5244 

2.1630 

4.0 

0.4951  d                      0.5356 

2.1424 

3.886 

0.5000  d 

0.5478 

2.1186 

3.75 

0.5045  d 

05593 

2.0974 

3.625 

0.5095  d 

0.5722 

2.0739 

3.5 

0.5146  d 

0.5855 

2.0493 

3.375 

0.5200  d 

0.5996                        2.0237 

3.25 

0.5256  d 

0  6145                        1.9970 

3.125 

0  5314  d 

0.6302                        1.9691 

3.0 

0.5375  d 

06468                        1.9399 

CIRCULAR   CAST-IRQ]*   BEAMS.  81 

From  this  Table  the  position  of  the  neutral  line  is  obtained 
by  placing  for  d,  the  diameter  in  inches 
and  computing  the  product  indicated. 

78.  The  Transverse  Strength 

of  Circular  Cast-Iron  Beams  may  be  ac- 
curately computed  from  either  Eq.  74-  or 
75,  but  as  this  involves  a  tedious  calcu- 
lation  Eq.  76  or  77  will  give  results,  practically  exact, 
much  less  labor. 

EXAMPLE  11. — Required  the  centre  breaking  load  of  a 
cular  Cast-Iron  Beam,  when 

The  diameter  d  =    2".0,  C  =  95200  pounds  computed,  Ex.  2, 

"     span s  —  20".0,  T  =  26569       "        by  test, 

m=4,          q  =  3.58. 

The  values  of  the  factors  fc  andyT,  obtained  by  interpolation 
from  the  Table,  page  80,  and  substituted  in  Eqs.  76  and  77, 
give 

T      4  (I)3  0.577  X  95200       1AOQfl 
L  —     v  ; — =  10986  pounds. 

20 

Major  "Wade,  in  his  experiments,  broke  this  beam  with 
11112  pounds. 

EXAMPLE  12. — Required  the  centre  breaking  load  of  a  Cir- 
cular Cast-Iron  Beam  when 

The  diameter.,  .d  —     2".42,  C  =  95200  Ibs.  computed,  Ex.  2, 

"     span s  =  20".0,    T  =  26569    "    by  test, 

m  =  4,  q  —  3.58. 

/T.  =  2.065  from  the  Table,  which,  substituted  in  Eq.  77,  gives 

26569 


82  STRENGTH   OF   BEAMS   AND    COLUMNS. 

Major  Wade  broke  four  beams  of  the  above  dimensions  with 
18141,  20419,  1999T  and  18225  pounds  respectively,  the 
mean  strength  being  19198  pounds. 

EXAMPLE  13. — Required  the  centre  transverse  elastic  limit 
load  of  a  Circular  Cast-Iron  Beam,  when 

The  diameter .  .d  —     T.129,  C  =  20000  pounds  by  test, 

"     span s  =  20".0,      T  =  17000       " 

m  =  4,  q  =  1.111. 

The  value  of  /T  =  1.1818  from  the  Table,  page  80,  substi- 
tuted in  Eq.  77,  gives 

T      4  (0.5645)'  1.1818  X  17000 
L--  -  =  1062  pounds. 

2i() 

The  elastic  limit  load,  of  the  two  of  these  beams  that  were 
tested,  was  1130  pounds  each,  the  values  of  C  and  T  being  the 
mean  of  three  tests,  as  given  in  the  United  States  Government 
Eeport  of  the  Tests  of  Iron  and  Steel  for  1884.* 

EXAMPLE  14. — Required  the  centre  breaking  load  of  the 
Circular  Cast-Iron  Beam  whose  elastic  limit  load  was  com- 
puted in  Example  13,  when 

C  =  100700  pounds  by  test,  /T  =  2.0333  from  the  Table, 
T=     29400        "  "          q-=  3.422. 

From  Eq.  77  we  have 

r       4  (0.5645)3  2.0333  X  29400 
L  -  — — —  -  =  2131  pounds. 

The  breaking  load  of  the  two  beams  described  in  Example 
13  was  2118  and  1795  pounds  respectively. 

79.  Movement  of  the  Neutral  Line.  In  the  Cir- 
cular Cast-Iron  Beam,  Example  13,  the  depth  of  the  neutral 

*  Senate  Ex.  Doc.  No.  35  -49th  Congress,  1st  Session,  p.  284. 


CIECULAE   CAST-IKON    BEAMS.  83 

line  below  tlie  compressed  surface  at  the  elastic  limit  was, 
from  the  Table,  page  80,  dc  =  OV7853  in  Example  14,  at  the 
instant  of  rupture,  dc  =  0".5846  ;  hence,  as  the  loading  pro- 
gressed, the  neutral  line  moved  upward,  or  from  the  tension 
side  toward  the  compressed  side  of  the  beam. 

8O.  Relative  Strength  of  Circular  and  Square 
Cast-iron  Beams. 

CASE  I. —  "When  the  circle  is  inscribed  within  the  square. 

The  relation  required  is  given  in  Eq.  83. 

EXAMPLE  15. — Required  the  relation  between  the  transverse 
strength  of  a  square  and  the  inscribed  circular  cast-iron  beam 
when  C  -r-  T  =  5. 

/  =  0.5  from  Table,  page  71,/c  =  0.4584  from  Table,  page  80. 

3X/C_3X0.4584 
8/a       '  8  X  (0.5)'   " 

.  • .  Strength  of  Circle  —  Strength  of  the  Square  X  0.6876. 

EXAMPLE  16. — Required  the  centre  breaking  transverse  load 
of  the  circular  beam  inscribed  within  the  square,  from  the 
tested  strength  of  the  square  beam,  when 

The  side  of  the  square  d  =  2".01,  q  =  3.58, 
"  diameter  of  the  circle  d=  2".01,/c  =  0.577  from  the  Table, 
"span a  =  20".0,/  =  0.561     «       " 

With  these  values  Eq.  83  becomes 

3  X/e  =  3  X  0.577  ? 

8  X  /*       8  (0.561)*  ~ 

Actual  b'k'g  load  square  beam,  Maj.  Wade's  tests,  16172  Ibs. 

"         "         "    circular   "         "  "       11112    " 

Comp'd   "  «         "       =16172X0.687  =  11110    " 

CASE  II. —  When  the  square  is  inscribed  within  the  circle. 

The  relation  will  be  given  by  Eq.  84. 

EXAMPLE  17. — Required  the  relation  between  the  strength 


84        STRENGTH  OF  BEAMS  AND  COLUMNS. 

of  the  circular  and  its  inscribed  square  beam,  when  C  -f-  jT=  5. 
/  =  0.5  from  the  Table,  page  Tl,/c  =  0.4584  from  the  Table, 
page  80. 

.          3/c        =3  X  0.4584 
'  2.8271  /a      2.8271  (0.5)2 

.  • .  Strength  of  the  Circle  =  /Strength  of  the  Square  X  1.945. 

CASE  III. —  When  the  circular  is  equal  in  area  to  that  of 
the  square  beam. 

The  relation  will  be  given  by  Eq..  85. 

EXAMPLE  18. — Required  the  relation  between  the  transverse 
strength  of  the  circular  and  the  square  beam,  when  their  areas 
are  equal  and  C  -f-  T  =  5. 

/  =  0.5  from  the  Table,  page  71, /c  =  0.4584  from  the  Table, 
page  80. 

.        3/e  3X0.4584 

'  5.564/2    ~  5.564  X  (0.5)2 

.  ' .  Strength  of  the  Circle  =  Strength  of  the  Square  X  0.987. 

EXAMPLE  19. — Required  the  centre  breaking  load  of  a  Circu- 
lar Cast-iron  Beam  from  that  of  the  square  beam  of  equal  area, 
when 

The  side  of  the  square d=    1".01,     q  —  5.091, 

' '    diameter  of  the  circle  d  =    1".  145,  /c  =  0.4562,  from  the  Table,  page  80, 
"    span *  =  60".0       /=  0.496      "       "        "        "     71. 


With  these  values  Eq.  85  becomes 
3  X  0.4562 


5.564  (0.496)5 


—  1. 


Mean  b'k'g  I'd  of  4  of  these  square  beams  from  Mr.  Barlow's  tests,*  519  Ibs., 

"          "     "  of  the  circular  beam  from  Mr.  Barlow's  tests,  519  pounds, 
Comp'd  "     "        "          "          "       =  519  X  1  =  519  pounds. 


*  Barlow's  "  Strength  of  Materials,"  p.  152. 


HOLLOW    CIRCULAR   CAST-IRON   BEAMS. 


85 


SECTION  VI. — Hollow  Circular  Cast-Iron  Beams. 
81.  Neutral  Lliie. 


Ratio  of  Crushing 
to  Tenacity,  or 
C  •*-  T  =  q. 

Depth  of  Neutral 
Line  Below  the 
Crushed  Side  of  the 
Beam,  or  dc- 

FACTORS  FOR  COMPUTING  THE  MOMENT 

OF  RESISTANCE,  r  =  1. 

/c  for  Crushing 
Strain,  C. 

/T  for  Tensile 
Strain,  T. 

8.319 

0.5000  d 

0.8584 

7.1414 

8.25 

0.5020  d 

0.8640 

7.1282 

8.125 

0.5057  d 

0.8742 

7.1104 

8.0 

0.5093  d 

0.8852 

7.1820 

7.875 

0.5131  d 

0.8962 

7.0580 

7.75 

0.5170  d 

0.9076 

7.0352 

7.625 

0.5209  d 

0.9190 

7.0078 

7.5 

0.5249  d 

0.9304 

6.9818 

7.375 

0.5290  d 

0.9430 

6.9550 

7.25 

0.5331  d 

0.9556 

6.9278 

7.125 

0.5374  d 

0.9684 

6.8996 

7.0 

0.5418  d 

0.9816 

6.8704 

6.875 

0.5462  d 

0.9950 

6.8410 

6.75 

0.5507  d 

1.0090 

6.8104 

6.625 

0.5553  d 

1.0232 

6.7790 

6.5 

0.5600  d 

1.0380 

6.7470 

6.375 

0.5647  d 

.0534 

6.7174 

6.25 

0.5697  d 

.0688 

6.6800 

6.125 

0.5748  d 

.0848 

6.6442 

6.0 

0.5800  d 

.1008 

6.6092 

5.875 

0.5852  d 

.1184 

6.5694 

5  75 

0.5907  d 

.1356 

6.5296 

5.625 

0.5962  d 

.1540 

6.4918 

5.5 

0.6018  d 

.1730 

6.4492 

5.375 

0.6076  d 

.1924 

6.4062 

5.25 

0.6134  d 

.2124 

6.3640 

5.125 

0.6193  d 

1.2330 

6.3184 

5.0 

0.6255  d 

1.2542 

6.2702 

4.875 

0.6314  d 

1.2748 

6.2266 

4.75 

0.6384  d 

1.2994 

6.1718 

4.625 

0.6451  d 

1.3230 

6.1192 

4.5 

0.6520  d 

*          1.3482 

6.0638 

4.375 

0.6590  d 

1.3740 

6.0064 

4.25 

0.6662  d 

1.4000 

5.9492 

4.125 

0.6735  d 

1.4274 

5.8877 

4.0 

0.6811  d 

1.4559 

5.8240 

3.875 

0.6889  d 

1.4856 

5.7569 

3.75 

0.6970  d 

1.5168 

5.6890 

3.625 

0.7052  d 

1.5488 

5.6148 

3.5 

0.7137  d 

1.5826 

5.5387 

3.375 

0.7223  d 

1.6172 

5.4578 

3.25 

0.7313  d 

1.6538 

5.3745 

3.125 

0.7405  d 

1.6922 

5.2880 

3.0 

0.7500  d 

1.7320 

5.1961 

86         STRENGTH  OF  BEAMS  AND  COLUMNS. 

The  above  Table  gives  the  positions  of  the  neutral  line  in  Hol- 
low Circular  Cast-Iron  Beams,  and  factors  for  use  in  Eqs.  91 
and  92. 

82.  Transverse  Strength.     The  transverse  strength 
^^ of  Hollow  Cast-Iron  Beams  may  be  ac- 

curately computed  from  either  Eq.  89 
or  90,  also  from  Eq.  91  or  92,  and  with 
much  less  labor. 

EXAMPLE  20. — Required  the  centre 
breaking  load  of  a  Hollow  Circular  Cast- 
iron  Beam,  when 

The  outer  diameter. . . .  =  3".875,  C  =  84000  pounds, 
"    inner       "         . . . .  =  3M25,  T,  =  14000        " 
"    mean       "         . .  d  =  3".5,       *,  =  6  feet, 
"    thickness  of  metal  t  =  0".375,/c,  =  1.1008  from  the  Table. 

By  using  these  values  in  Eq.  91  and  for  r  =  1.75,  one  half 
of  the  mean  diameter  the  load,  Z,  becomes 

L  =  4X0.375(1.75)- 1.1008X84000  =  5g99 
12  X  6 

The  actual  breaking  load  was  5122  pounds  from  the  series  of 
experiments  described  in  Example  10. 


CHAPTEE  V. 

WROUGHT-IRON   AND    STEEL   BEAMS. 
SECTION  I. — General  Conditions — Wrought-iron. 

83.  Compressive  Strength.      Crushing. — Wrought- 
iron,  when  subjected  to  pressure,  increases  its  area  so  rapidly 
that  it  is  impossible  to  determine  with  precision  its  crushing 
strength  or  that  intensity  of  pressure  that  corresponds  to  the 
tenacity.     Its  value,  per  square  inch,  as  quoted  by  the  various 
writers,  varies  from  30000  to  90000  pounds  per  square  inch. 

The  great  pressure  to  which  wrought-iron  is  subjected  when 
rolled  into  sheets  and  eye-beams  causes  it  to  lose  a  portion  of 
its  ductility  and  increase  its  crushing  strength.  The  crushing 
strength  of  Swedish  bar-iron,  computed  from  Mr.  Kirkaldy's 
experiments  (Example  21),  is  60000  pounds  per  square  inch, 
while  the  crushing  strength  of  wrought-iron  when  rolled  into 
eye-beams,  computed  from  Mr.  Fairbairn's  experiments,  is 
86466  pounds  per  square  inch  (Example  22). 

Elastic  Limit. — The  compressive  elastic  limit  of  wrought- 
iron  in  bars  may  be  taken  at  30000  pounds  per  square  inch. 
The  computed  value  of  the  compressive  elastic  limit  for  rolled 
eye-beams  is  631  TO  pounds  (Example  31),  from  data  furnished 
by  the  Phoenix  Iron  Company's  experiments. 

84.  Tensile  Strength.   Tenacity. — When  wrought-iron 
is  subjected  to  a  tensile  strain  before  rupture  takes  place,  it  in- 
creases in  length  from  15  to  20  per  cent,  and  contracts  in  area 
at  the  fractured  section  about  25  per  cent.     Its  tenacity  ranges 
in  value  from  50000  to  65000  pounds  per  square  inch  ;  wrought- 
iron  diminishes  in  tensile  strength  when  rolled  into  sheets. 


88  STRENGTH    OF   BEAMS   AND    COLUMNS. 

Elastic  Limit. — The  tensile  elastic  limit  of  bar-iron  is  about 
one  half  of  its  tenacity ;  the  mean  may  be  considered  to  be 
30000  pounds  per  square  inch. 

STEEL. 

85.  Compress! ve   Strength.     Crushing. — In   deter- 
mining its  crushing  strength,  the  same  difficulty  mentioned  in 
determining  that  of  wrought-iron  is  encountered  ;  its  value,  as 
given  for  the  different  kinds,  varies  from  100000  to  340000 
pounds  per  square  inch. 

The  compressive  elastic  limit  of  steel  ranges  from  20000  to 
60000  pounds  per  square  inch. 

86.  Tensile  Strength.     The  tenacity  of  .steel  is  greater 
than  that  of  any  known  material ;  it  ranges  in  value  from 
60000  to  160000  pounds  per  square  inch. 

87.  To  Compute  the  Compress! ve   Strength. 

Formula  31,  page  40,  and  Eq.  79,  page  58,  may  be  used  to  great 
advantage  to  determine  the  crushing  strength  for  such  materials 
as  wrought-iron  and  steel  whose  resistance  to  crushing  cannot 
be  readily  determined  in  the  usual  manner,  because  they  in- 
crease their  area  so  rapidly,  under  direct  pressure,  when  ap- 
plied to  small  test  specimens,  that  no  precise  determination  of 
their  strength  can  be  made.  The  deflection  of  the  beam,  how- 
ever, must  be  sufficient  to  allow  the  neutral  line  to  move  to  the 
position  required  for  rupture  of  the  fibres  by  the  cornpressive 
and  tensile  strains  at  the  same  instant,  for  the  reason  given  in 
Art.  24. 

EXAMPLE  21. — Required  the  Crushing  Value  of  C  for 
wrought-iron,  from  the  tensile  strength  and  the  centre  trans- 
verse breaking  down  load,  when 

The  depth d  =    2.0  inches,  T=  42133  pounds,  by  test, 

"     breadth...  &  =    2.0       "         L  =  13338       "         "      " 
"     span s  =  25.0       "         m  =  4. 


GENERAL   CONDITIONS.  89 

The  required  position  of  the  neutral  line  from  Eq.  30  is 


j  —  3  X  2     ^9X4X2  (2)2  42133- 12  X  13338  X25_ 
T  ~       2  4  X  4  X  2  X  42133"  °'554' 

.  • .  dc  =  2  -  0.554  =  1.446  inches  and  C  from  Eq.  31. 

^_3  X  13338  X  25  _ 
4  X  2(1.446)* 

This  example  is  taken  from  Mr.  Kirkaldy's  experiments.* 
By  direct  .pressure  he  determined  C  —  84890  pounds,  when 
the  length  =  2  diameters,  and  0=  148840  pounds,  when  the 
length  —  1  diameter  of  the  test  specimen.  Our  computation  of 
the  value  of  C  is  made  under  the  hypothesis  that  the  tensile 
strain  in  the  experiment  was  sufficiently  intense,  though  it  did 
not  actually  fracture  the  beam. 

EXAMPLE  22. — Kequired  the  Crushing  Yalue  of  C  for  rolled 
wrought-iron  from  the  tensile  strength  and  the  centre  trans- 
verse breaking  load  of  a  rolled  wrought-iron  T  beam. 

The  depth  of  the  beam  d  —  3".0,  T  —  57600  pounds  mean 
for  British  bar-iron. 

The  breadth  of  the  web  b  =  0".5,  L  =  2690  pounds,  from 
Mr.  Fairbairn's  test.f 

The  breadth  of  the  compressed  flanges  &,  —  2".5,  m  =  4. 
The  deptli  "         d,  =  0".375,  a  =  10  ft. 

The  position  of  the  neutral  line,  the  beams  having  no  bottom 
flanges,  from  Eq.  30,  becomes 

d,  =  3  X  8  _  .4 /9~X  4  X  0.5  (3)2  57600  -  12  X  2690  X  120  _  .,  Q6  . 
V  4  X  4  X  0.5  X  57600 


*  Barlow's  "Strength  of  Materials,"  p.  256. 
t  Box's  •'  Strength  of  Materials,"  p.  214. 


90  STRENGTH   OF   BEAMS    AND   COLUMNS. 

arid  from  Eq.  37, 
L  =  2690  =  0.5  (04)3  + 


EXAMPLE  23.  —  Required  the  Crushing  Strength  of  certain 
steel  made  bj  the  Otis  Iron  and  Steel  Co.,  of  Cleveland, 
Ohio,  from  the  centre  breaking  load  of  a  circular  beam  when 

The  diameter  =  1".129,  T  —  83500  pounds  mean  of  10  tests, 
"    span         —  20".0,     Z  =  3860         "       one  test. 


From  Eq.  78  we  have 


*  ~~  4  (0.5645)3  83500 

From  the  Table,  page  100,  q  —  1.2122  for  this  value  of  /T, 
and  from  Eq.  79, 

C  =  1.2122  X  83500  —  101218  pounds. 

This  example  is  taken  from  the  "United  States  Government 
Report  of  Tests  of  Iron  and  Steel,  at  Watertown  Arsenal,  for 

1885.* 

88.   To   Compute  the   Tensile   Strength.      The 

tensile  strength  may  be  computed  from  test  values  of  C  and 
the  load  L  in  rectangular  beams  by  means  of  Eq.  33,  having 
first  ascertained  the  position  of  the  neutral  line  from  Eq.  32, 
and  for  circular  beams  from  Eqs.  80  and  81,  page  58. 

EXAMPLE  24.  —  Required  the  Tensile  Strength  of  "  Burden's 
Best  "  wrought-iron  from  the  centre  breaking  load  of  a  cir- 
cular beam  when 

*  Senate  Ex.  Doc.  No.  36—  49th  Congress,  1st  Session,  p.  690. 


GENERAL   CONDITIONS.  91 

The  diameter  .......  d  =  1".25,    C  —  64500  pounds  assumed, 

"    span  ...........  *  =  12".0,  Z  =    6000       "       by  test. 


From  Eq.  80  we  have 


6000  X  12        _ 
' 


c  ~~  4  (0.625)'  64500 

From  the  Table,  page  100,  q  =  1  for  the  above  value  of  yc, 
from  Eq.  81, 

T  =  ^>2?  =  64500  pounds. 

This  beam  was  broken  at  the  Rensselaer  Polytechnic  In- 
stitute in  November,  1882,  as  given  by  Professor  W.  H.  Burr. 

89.   Transverse    Strength    of   Wrought  -  Iron. 

The  elastic  limit  load  is  technically  called  the  breaking  load 
for  wrought-iron  beams,  as  from  its  position  in  the  section  of 
the  beam  where  fracture  should  take  place  from  the  fibre 
strains  being  greatest,  it  cannot  undergo  the  change  of  form 
required  before  its  fibres  can  be  fractured  /  this  breaking  load 
may  be  computed  from  the  tensile  and  compressive  elastic 
limit  coefficients  of  wrought-iron. 

The  breaking-down  load.  —  While  wrought-iron  in  beams 
cannot  be  broken  transversely  by  rupturing  its  fibres,  yet  there 
is  an  intensity  of  the  transverse  lo*ad  at  which  it  does  not 
appear  to  be  able  to  offer  any  further  resistance  to  the  action 
of  the  bending  load;  this  load  may  be  called  the  breaking- 
down  load  ;  in  wrought-iron  beams  it  is  about  twice  the 
technical  breaking  or  elastic  limit  load. 

Wrought-iron  when  rolled  into  T  beams  may  become 
sufficiently  hard  and  unyielding  from  the  intense  pressure  re- 
quired to  make  it  fill  the  rolls,  that  it  will  offer  sufficient 
resistance  to  actually  fracture  the  fibres  by  the  tensile  strain,  as 
in  Example  22. 


92        STRENGTH  OF  BEAMS  AND  COLUMNS. 

9O.   Transverse   Strength   of  Steel.      As   in  the 

case  of    wrought-iron  beams  the  elastic  limit  load    of   steel 
beams  is  technically  called  the  breaking  load. 

The  breaking-down  load  of  steel  beams  is  about  fths  of  the 
elastic  limit  load. 


SECTION  II. — Rectangular  Wrought-iron  and  Steel  Beams. 

91.  Neutral  Line.  The  position  of  the  neutral  line 
may  be  computed  from  either  formula  25  or  26.  From  Eq. 
25  the  position  has  been  computed  for  the  different  values  of 
(7  -r-  T  —  q,  that  are  required  in  wrought-iron  and  steel  rect- 
angular beams,  and  the  results  tabulated  below  for  reference. 

Table  of  Positions  of  the  Neutral  Line  in  Rectangular 
Wro^lght-Iron  and  Steel  Seams  : 


Ratio  of  Crushing 
to  Tenacity,  or 
C  -*-  T  =  q. 

Depth  of  Neutral  Line 
Below  the  Crashed 
Side  of  the  Beam, 
or  rfc. 

Ratio  of  Crushing 
to  Tenacity,  or 
C  •*•  T  =  q. 

Depth  of  Neutral  Line 
Below  the  Crushed 
Side  of  the  Beam, 
or  rfc. 

3.0 

0.5930  d 

1.9 

0.6757  d 

2.9 

0.5993  d 

1.8 

0.6852  d 

2.8 

0.6057  d 

1.7 

0.6951  d 

2.7 

0.6124  d 

1.6 

0.7056  d 

2.6 

0.6193  d 

1.5 

0.7161  d 

2.5 

0.6264  d 

1.4 

0.7280  d 

2.4 

0.6338  tt 

1.3 

0.7401  d 

2.3 

0.6416  d 

1.2 

0.7529  d 

2.2 

0.6496  d 

1.1 

0.7664  d 

2.1 

0.6579  d 

1.0 

0.7807  d 

2.0 

0.6666  d 

0.9 

0.7960  d 

For  ratios  greater  than  C  -T-  T  =  3  refer  to  the  Table,  page 

80. 

92.   Transverse  Strength.     Either  Eq.  27  or  28  may 

be   used   to  compute   the  transverse  strength  of  rectangular 
wrought-iron  and  steel  beams,  as  may  be  most  convenient. 


RECTANGULAR   WROUGHT-IRON   AND   STEEL   BEAMS.    93 

EXAMPLE  25. — Required  the  centre  breaking  or  elastic  limit 
load  of  a  Rectangular  Wrought-Iron  Beam,  when 

The  depth d  =     3".0,    C  =  30000  pounds  mean  value, 

f  "    breadth I  =     1".5,   T  =  30000       "  " 

'  "    span s  =  33".0,    q  =  1,  and  m  =.  4. 

The  position  of  the  neutral  line  may  be  computed  or  taken 
directly  from  the  Table  f or  q  =  1. 

d-0  =  0.7805  X  3  =  2".3415,  dT  =  3  —  2.3415  =  0.6585. 

From  Eq.  28  the  value  of  the  load,  L,  required  becomes 

T      4  X  1.5  X  0.6585  (2x3  +  2.3415)  30000 

L  —  -  — ^ ^r— -  -  =  9681  pounds. 

o  X  £>*> 

Mr.  Barlow  tested  this  iron  and  found  its  tensile  elasticity 
perfect  with  22400  pounds  ;  the  transverse 
elastic  limit  load  was,  by  experiment,  be- 
tween 9520  and  10080  pounds.  A  num- 
ber of  beams  tested  gave  similar  re- 
sults. 

The  deflection  with  10080  pounds  was 
0".963  in  one  bar  and  0".624  in  another ; 
hence  the  elastic  fibre  strain  limits  were  fully  developed.* 

EXAMPLE  26. — Required  the  centre  breaking  or  elastic  limit 
load  of  a  Rectangular  Wrought-Iron  Beam  of  Swedish  iron, 
when 

The  depth d  =     2" .0,   C  **  22637  pounds  by  test, 

"     breadth b=    2".0,  T  =  24052       " 

"      span s=  25".0,    q  —  0.94,  and  m  —  4. 

The  position  of  the  neutral  line  may  be  computed  from  Eq. 
25,  or  taken  from  the  Table,  by  proportion  between  the  posi- 


*  Barlow's  "  Strength  of  Materials,"  p.  278. 


94         STRENGTH  OF  BEAMS  AND  COLUMNS. 

tions  for   q  =  1  and    q  =  0.9,  for  which  dc  =  1.58.     With 
these  values  the  required  load,  Z,  becomes  from  Eq.  27, 

T       4  X  2(1.58)222631 

=  ~ 3X25  r  =  P°imds- 

Mr.  Kirkaldy  determined,  experimentally,  that  the  centre 
breaking  load  of  this  beam  was  between  6000  and  6500 
pounds,  and  that  the  elastic  limit  values  of  O  and  T  were  as 
given  above. 

The  position  of  the  neutral  line  for  the  l>reaking-down 
load  of  this  beam  (Example  21)  was  1".446  below  the  top  or 
compressed  side  of  the  beam,  and  as  the  loading  progressed 
it  must  have  moved  upward  from  its  position  at  the  elastic 
limit,  1".58,  to  1".446,  its  position  with  the  breaking-down 
load. 

The  deflection  with  6000  pounds  was  0".318,  and  with  6500 
0".506  ;  it  only  required  0".46  to  fully  develop  the  elastic 
fibre  strain  limits.* 

EXAMPLE  21. — Required  the  centre  transverse  elastic  limit 
load  of  a  Rectangular  Steel  Beam,  when 

The  depth d=     1".15,  C  =  48000  pounds  by  test, 

"     breadth I  =     1".75,  C  =  52000       " 

"     span s  =  25".0,      q  =  0.923,  and  m  —  4. 

The  position  of  the  neutral  line,  computed  from  Eq.  25,  or 
taken  from  the  Table  by  proportion,  is 

dc  =  1".3S67  and  </T  =  0.3633. 
The  required  load,  Z,  from  Eq.  21,  becomes 

T       4  X  1.15  (1.3867)*  48000 

=  ~"3  X25  pounds. 

• 

*  Barlow's  "  Strength  of  Materials,"  p.  256. 


RECTANGULAR    WROUGHT-IRON   AND    STEEL   BEAMS.    95 

Mr.  Kirkaldy  determined  by  experiment  that  the  elastic 
limit  load  of  each  of  4  of  these  beams  was  between  8000  and 
9000  pounds,  and  that  the  values  of  C  and  T  were  as 
given. 

The  deflection  with  8000  pounds  was  0".476,  and  it  only 
required  0".3633  to  develop  the  elastic  compressive  and  tensile 
fibre  strain  limits  at  the  same  instant.* 

EXAMPLE  28. — ^Required  the  breaking-down  load  of  the 
beam  described  in  Example  27,  when 

C  =  159582  pounds  by  test,  q  =  2.3, 
T  =     69336       «  "       m  =  4. 

For  the  position  of  the  neutral  line  from  Eq.  25  or  from  the 
Table, 

d0  =  1".121  and  d,  =  0".629. 

The  breaking-down  load,  .Z,  becomes  from  Eq.  27, 

T       4  X  1.75  (1.121)2  159582 

L  =  -          — ~ •—-  -  =  18716  pounds. 

o  X  -aO 

The  mean  breaking-down  load  of  4  cf  these  beams  was 
16477  pounds  from  Mr.  Kirkaldy's  experiments.  The  upward 
movement  of  the  neutral  line  is  seen  in  these  Examples,  in 
Example  27,  dc  =  l//.3867  and  in  Example  28,  dc  =  1".121. 


Barlow's  "  Strength  of  Materials,"  p.  253. 


96        STRENGTH  OF  BEAMS  AND  COLUMNS. 


SECTION    III. — Double    T,  Rolled    Eye-Beams   and  Hollow 
Rectangle  or  Box,  Wrought-Iron  and  Steel  Beams. 

93.  Neutral  Line.    The  neutral  lines  in  these  wrought- 
iron  and  steel  beams  are  so  near  the  bottom  or  tension  side  of 
the  beam  that  their  positions  cannot  be  computed  from  ap- 
proximate formulas,  and  the  exact  formula  is  too  complicated 
to  be  useful  in  practice.     Before  beams  of  these  sections  are 
manufactured,  its  position  should    be  assumed  and  a  sufficient 
area  of  the  metal  placed  above  and  below  the  assumed  neutral 
line  to  insure  equilibrium  from  the  known  values  of  the  crush- 
ing and  tensile  strength  of  the  material. 

In  most  practical  examples  of  these  beams  the  neutral  line 
is  located  within  the  tension  flange,  and  its  position  can  be 
computed  from  Eq.  30,  page  40,  wrhen  L  the  load  and  T  the 
tensile  strength  have  been  determined  experimentally,  or  by 
the  methods  given  in  Problems  I.  and  II.,  page  43. 

94.  Transverse   Strength .      When  the  position  of 
the  neutral  line  is  known  the  transverse  strength  may  be  com- 
puted from   either  Eq.  37  or  38,   that  were  deduced  for  the 
transverse  strength  of  the  Hodgkinson  beam,  giving  the  letters 
of  the  formulas  the  meaning  defined  in  Art.  44,  page  47. 

The  transverse  strength  of  both  the  Tee  and  Double- 
Headed  Railroad  Rails  may  be  very  accurately  computed  from 
the  above  formulas,  although  they  are  partly  bounded  in 
outline  by  curved  lines.  The  web  must  extend  from  the 
bottom  to  the  top  of  the  rail,  and  an  equivalent  in  area  right 
line  section  must  be  formed  for  the  metal  that  remains  in  both 
the  base  and  head  of  the  rail,  which  will  enable  the  formulas 
to  be  applied  to  these  forms  of  beams. 

95.  Double    T     beams   loith  tension  and   compression 
jlanges  unequal  in  area. 

EXAMPLE  29. — Required  the  centre  elastic  limit  load  of  a 
Wrought-Iron  Rolled  Double  T  Beam,  when 


DOUBLE   T,    EOLLED   EYE-BEAMS,    ETC.  97 

Depth  of  beam :d   =  8".  38,  C  =  36320  Ibs.  required  for  eq'brium, 

Breadth  of  web b   =  0".325,  T  =  30000  pounds  mean  value,' 

"  top  flanges  bl  =  2".  175,  ra  =  4, 

Depth     "     "        "       di  =  1".0,  s   =  11  feet, 

"  bottom  "      dz  =  0".38, 
Breadth  "       "       "       Ja  =  3".  675. 

The  neutral  line  will  be  assumed  to  be  

near  the  top  line  of  the  tension  flanges, 
or  d, . 

dc  =  8/r.05  and  d,f  =  0.33.  I 


The  elastic  limit  load,  Z,  becomes  from 


Eq.  37, 


_  0.325(8.05)3  +  2.175  <!)»  (3  X  8.05- 2X1)  _ 

3~X~a05  X  132  yy/<54  lbs> ' 

and  from  Eq.  36, 

T       4  X  4  (3  X  8.38  +  8.05)  30000 

3X132  ^9924  pounds. 

The  elastic  limit  load  was  between  the  applied  loads  9493 
and  11253  pounds  from  Mr.  William  Fairbairn's  experiments, 
and  their  deflections  were  0".46  and  0".60  ;  the  deflection  re- 
quired was  0".33.* 

EXAMPLE  30. — Required  the  centre  elastic  limit  transverse 
load  of  a  Rolled  Wrought-Iron  Double  T  Beam  when  ' 

Depth  otT  beam d  =  9". 44,    C  =  37165  Ibs.  required  for  eq'brium, 

Breadth  of  web b  =  0".35,  T  =  30000  pounds  mean  value, 

"  top     flanges  bl  —  2". 4,      *  =  10  feet, 
Depth      "     "  "     di  -  1".0,     m=4, 

"  bottom     "     da  =  0".44,  dc  =  9.06  assumed, 
Breadth"      "          ••      &2  =  3". 95,  dT  =  0.38        " 


*  Fairbairn,  "  On  Cast  and  Wrought-Iron,"  p.  102. 


98        STRENGTH  OF  BEAMS  AND  COLUMNS. 

With  these  values  the  load,  Z,  becomes  from  Eq.  28, 

T       4  X  4.3  X  0.38  (2  X  9.44  +  9.06)  30000 
L  =  3  X  12  X  1<T-  ^ 

and  from  Eq.  38, 

0.35(9.06)3  +  2.4  (I)2  (3  X  9.06  -  2  X  1)       ,       _„,  _.  ._  .. 

~¥^9766~xi2x~io~~ 

The  elastic  limit  load  was  between  the  applied  loads  14693 
and  16373  pounds  from  Mr.  William  Fairbairn' s  experiments, 
and  the  deflection  was  0".35  and  0".45  with  these  loads  re- 
spectively.* 

In  Examples  29  and  30  the  tensile  elastic  limit  has  been 

assumed  to  be  30000  pounds,  the  mean  value   for  wrought- 

iron,  and  from  it  the  position  of  the  neutral  line  has  been 

computed  from  Eq.  30,.  and  then  the  required  value  of  C  to 

produce  equilibrium. 

rr 


96.    Double  T  'beams  with  tension 
c  T>  j     an^  compression  flanges  equal  in  area. 


*--- 
**±^ 


~     ~ 


1. 


EXAMPLE  31. — Required  the  centre 
transverse  elastic  limit  load  of  a  Rolled 
Wrought-Iron  Eyebeam  when 


Depth  of  beam d  =  9".0,     0  —  63170  Ibs.  required  for  eq'brium, 

Breadth  of  web b   =  0".6,     ^=30000    "    mean  value, 

"        "  top  flanges  &i  =  4". 775,  ra  =  4, 
Depth      "     "        "      di  =  1".0,      s=  14  feet, 

"  bottom"     d.2  =  1".0,      dc  =  8".2, 
Breadth  "       "       "      5a  =  4".775,  dT  =  0".8. 

With  these  values  the  load,  Z,  becomes  from  Eq.  37, 

L  =  ^^xa-'xl)  >< 4  ><  -™ 


*  Fairbairn,   "  On  Cast  and  Wrought-Iron,"  p.  103. 


DOUBLE   T,    ROLLED   EYE-BEAMS,    ETC.  99 

and  from  Eq.  28,  in  which  b  =  I  +  \  =  5.375, 


The  value  of  T  is  taken  at  the  mean,  from  which  dv  must 
be  0".8  and  O=  63170,  that  equilibrium  shall  exist  between 
the  moments  of  the  tensile  and  compressive  resistances. 

The  rolled  eyebeam,  of  which  the  data  given  in  Example 
31  is  the  equivalent  right  line  section,  was  tested  by  the  Phoe- 
nix Iron  Co.  of  Pennsylvania,*  and  its'  centre  elastic  limit  load 
was  found  to  be  between  the  applied  loads  26880  pounds  and 
28000  pounds  ;  the  deflection  was  0//.5T2  and  0".600  with 
these  loads  respectively. 

The  proportion  of  the  load  and  area*  of  the  section  that  is 
sustained  by  each  member  is  given  in  the  following  table  : 

Load.  Area. 

Compression  Flanges  ..............  12.3  31.9  per  cent. 

"  Web  ................  37.7  32.9       " 

Tension  "     ................   0.55  3.2       " 

"       Flanges  ...................  49.45  24.1       " 

"  "     (practically  lost)  ......  7.9       " 


100.00          100.0 

The  transverse  elastic  limit  load  of  the  rectangular  wrought- 
iron  beam  5".375  X  9".0,  from  which  the  above  described 
eyebeam  may  be  supposed  to  have  been  cut,  is  from  Eq.  27, 
when  C  =  30000  and  T  =  30000  pounds,  the  mean  values 
for  bar-iron, 

T       4  (7.02)2  30000        ,on_ 
L  ="  3  X  12  X  14  :  ~-  6S076  P°Unds' 

From  which  it  will  be  observed  that  the  eyebeam,  while  con- 
taining only  31  per  cent  of  the  area  of  the  rectangular  beam, 

*  Phoenix  Iron  Company's  "  Handbook  of  Useful  Information." 


100 


STRENGTH   OF   BEAMS   AND    COLUMNS. 


is  able  to  sustain  42.5  per  cent  of  its  load,  which  is  supposed 
to  be  due  to  the  elevation  of  the  elastic  limit  during  the  pro- 
cess of  rolling,  or  the  top  of  the  beam  must  have  been  laid 
with  steel. 


SECTION  IY. — Circular  Wrought- Iron  and  Steel  Beams. 

97.  Neutral  Line.  The  position  of  the  neutral  line 
in  Circular  Wrought-Iton  and  Steel  Beams  for  the  different 
ratios  of  C  -f-  T  =  q  required  has  been  computed,  also  the 
factors  f/c  and/i  for  use  in  Eqs.  76  and  77,  and  tabulated  be- 
low for  reference. 


FACTORS  FOR  COMPUTING  THE  MOMENT  op 

Ratio  of  Crushing 
to  Tenacity,  or 
C  -*•  T  =  q. 

Depth  of  Neutral 
Line  Below  the 
Crushed  Side  of  the 
Beam,  or  dc. 

RESISTANCE,  RADIUS  =  1. 

/c  for  Crushing 

T  for  Tensile 

Strain  C. 

Strain  T. 

3.0 

0.5375  d 

0.6468 

1.9399 

2.875 

0.5438  d 

0.6642 

1.9093 

2.75 

0.5505  d 

0.6827 

1.8766 

2.625 

0.5574  d 

0.7024 

1.8434 

2.5 

0.5646  d 

0.7230 

1.8084 

2.375 

0.5724  d 

0.7458 

1.7718 

2,25 

0.5804  d 

0.7696 

1.7314 

2.125 

0.5890  d 

0.7952 

1.6896 

2.0 

0.5980  d 

0.8228 

1.6454 

1.875 

0.6076  d 

0.8526 

1.5984 

1.75 

0.6178  d 

0.8848 

1.5482 

.625 

0.6277  d 

0.9198 

1.4944 

.5 

0.6404  d 

0.9580 

1.4368 

.375 

0.6525  d 

1.0000 

1.3748 

.25 

0.6666  d 

1.0462 

1.3074 

.125 

0.6816  d 

1.0976 

1.2328 

.0 

0.6976  d 

1.1548 

1.1548 

From  this  Table  the  position  of  the  neutral  line  in  any 
wrought-iron  or  steel  beam  may  be  obtained  by  multiplying 
the  diameter  d,  expressed  in  inches,  by  the  decimal  factor 


CIRCULAR  WROUGHT-IRON   AND   STEEL   BEAMS.       101 

corresponding  to  the  ratio  C  -r-  T  =  q.     For  ratios  intermedi- 
ate in  value  the  position  may  be  obtained  by  proportion. 

98.   Transverse   Strength.     This  may  be  computed 

from  either  Eq.  74  or  75,  but  with  much 

less  labor  from  Eq.  76  or  77,  using  the 
values  of  fc  and  /*T  given  in  the  above 
Table. 

EXAMPLE  32. — Required  the  centre 
breaking  or  elastic  limit  load  of  a  Circu- 
lar Steel  Beam  when 

The  diameter  d  =  1".129,  C  =  41160  Ibs.  mean  of  four  tests, 

"    span s  =  20".0,     T  =  39200  "        "       "  ten       " 

"    factor...  m  =  4,  q  =  1.05. 

The  position  of  the  neutral  line  and  factor,  f&  becomes,  by 
proportion  from  the  Table, 

dc  =  0.6912  X  1".129  =  0.780,  /c  =  1.132. 
With  these  values  the  load,  L,  required  becomes  from  Eq.  76, 

T       4  (0.5645)*  1.132  x  41160 

L  —  -^—  -  =  1676  pounds. 

Z(J 

From  the  United  States  Government  Report  of  the  Tests 
of  Iron  and  Steel  for  the  year  1885,*  the  elastic  limit  load  of 
two  of  these  beams  that  were  tested  was  1638  pounds  with 
the  values  of  C  and  T  as  given  in  the  Example. 

EXAMPLE  33. — Required  the  centre  elastic  limit  load  of  a 
cylindrical  Phoenix  pin  supported  at  both  ends  when 

*  Senate  Ex.  Doc.  No.  36-49th  Congress,  1st  Session,  p.  690. 


102  STRENGTH    OF   BEAMS   AND    COLUMNS. 

The  diameter. ...    d=    2".5,  C  =    30000  pounds  from  tests, 

«    span s  =  24".0,  T  =    30000       " 

" '  factor m  =  4,         /c  =  1.1548  from  the  Table. 

Then  from  Eq.  76  we  have  for  the  required  load,  • 

L  -  JW  1.1S48  X  30000 
24 

The  elastic  limit  strength  of  this  pin  was  11000  pounds. 
(Watertown  Arsenal  Report  of  Tests  of  Iron  and  Steel  for 
1881.)*  The  computed  ultimate  strength  is  18795  pounds, 
with  0  and  T  =  50000  pounds,  the  recorded  ultimate  strength 
is  20000  pounds. 

In  order  that  these  wrought-iron  pins  should  truly  cross- 
break,  their  deflection  should  not  be  less  than  0.3024  d',  in  the 
above  examples  the  ultimate  deflection  was  l//.278  ;  while  that 
required  for  true  cross-breaking  was  0".75,  the  recorded  de- 
flection with  18000  pounds  was  0".78. 

A  number  of  these  pins,  made  of  Phoenix  and  Pencoyd 
iron,  were  tested  with  diameters  ranging  from  2J  to  5  inches, 
but  the  span  was  so  short,  in  nearly  all  cases,  that  the  tensile 
fibre  strain  was  not  fully  developed,  causing  the  transverse 
elastic  limit  load  to  be  not  well  defined  and  the  ultimate  load 
to  be  larger  than  it  should  be.  Though  the  computed  elastic  limit 
loads  do  not  differ  very  greatly  from  those  determined  by 
tests,  there  is  a  very  great  difference  between  the  computed 
and  experimental  ultimate  loads,  the  latter  being  the  greater. 

This  series  of  tests  illustrates  the  correctness  of  the  state- 
ment made  in  Art.  24,  that  the  bending  moment  of  the  ap- 
plied load  at  the  inception  of  the  deflection  of  a  beam  is  held 
in  equilibrium  by  the  moment  of  a  purely  compressive  resist- 
ance that  is  distributed  over  the  section  in  an  uniformly  vary- 

*  House  of  Representatives  Ex.  Doc.  No.  12,  1st  Session,  47th  Congress, 
p.  171. 


CIRCULAR   WBOUGHT-IRON   AND    STEEL   BEAMS.       103 

ing  strain.  Test  ISTo.  741,  page  186,  of  the  Report,  was  of 
Phoenix  iron  4"  in  diameter  and  24"  span.  The  bending 
moment  of  the  centre  applied  load  is  6  L  from  Eq.  3,  page  5, 
and  the  moment  of  compressive  resistance  is  the  product  of 
the  resultant  (the  area  of  the  section  by  one  half,  (7,  the  elastic 
limit  compressive  strength),  by  its  lever-arm,  f-6?,  the  distance 
of  the  centre  of  gravity  of  the  pressure  wedge  below  the  axis  ; 
hence 

3d  .         (V       C 


.  .  .  L  = 


The  observed  elastic  limit  load  was  48000  pounds  with  a 
deflection  of  0".0585,  but  as  no  correction  appears  to  have 
been  made  for  the  settling  of  the  beam  on  its  bearings,  we 
may  conclude  that  the  beam  was  on  the  verge  of  true  deflec- 
tion with  its  transverse  elastic  limit  load. 


104 


STRENGTH   OF   BEAMS  AND    COLUMNS. 


SECTION  Y ' .—Hollow   Circular  Wrought-Iron  and  Steel 

Beams. 

99.  Neutral  Line.  The  position  of  the  neutral  line  in 
Hollow  Circular  Wrought-Iron  and  Steel  Beams  for  the 
different  ratios  of  C  -j-  T  =  q  required  has  been  computed, 
also  the  factors  fG  and  fT  for  use  in  Eqs.  91  and  92,  and 
tabulated  below. 


FACTORS  FOR  COMPUTING  THE  MOMENT  OF 

Ratio  of  Crushing 
to  Tenacity,  or 

Depth  of  Neutral 
Line  Below  the 
Crushed  Side  of  the 

RESISTANCE,  RADIUS  =  1. 

C  -t-  T  =  q- 

Beam,  or  dc. 

/d"for  Crushing 
Strain  C. 

ft  for  Tensile 
Strain  T. 

3.0 

0.7500  d 

1.7320 

5.1961 

2.875 

0.7597  d 

1.7737 

5.0992 

2.75 

0.7697  d 

1.8172 

5.0000 

2.625 

0.7800  d 

1.8628 

4.8898 

2.5 

0.7906  d 

1.9105 

4.7760 

2.375 

0.8014  d 

1.9601 

4.6566 

2.25 

0,8122  d 

2.0108 

4.5334 

2.125 

0.8241  d 

2.0645 

4.3928 

2.0 

0.8358  d 

2.1251 

4.2490 

1.875 

0.8478  d 

2.1853 

4.0966 

1.75 

0.8600  d 

2.2479 

3.9338 

.625 

0.8727  d 

2.3135 

3.7586 

.5 

0.8851  d 

2.3823 

3.5722 

.375 

0.8979  d 

2.4534 

3.3728 

.25 

0.9107  d 

2.5267 

3.1588 

.125 

0.9233  d 

2.6024 

2.9296 

1.0 

0.9360  d 

2.6806 

2.6806 

This 


may 


be 


coiri- 


1OO.   Transverse     Strength. 

puted  from  either  Eq.  89,  90,  91  or  92. 

EXAMPLE  34. — Required  the  centre  breaking  transverse  load 
of  a    Hollow   Wrought-Iron  Cylindrical  Beam,  supported  at 
both  ends,  when 
The  outer  diameter. .  .d  =  •  4".0,  <7and  T  =  45000  pounds, 

"    thickness  of  metal  t  =  0".1875,     /c  =  2. 6806  from  Table, 

"    span s  =    6'.0  in  =  4. 


WROUGHT-IRON  AND   STEEL   BEAMS.  105 

Then  from  Eq.  91  we  have  for  the  load, 

T       4  (2)2  2.6806  x  0.1875  x  45000 
L  =  -  —t  -  —  5076  pounds. 

This  "  tube  failed  suddenly  with  5824  pounds."  (From  "  The 
Britannia  and  Conway  Tubular  Bridge,"  p.  435.) 


CHAPTER  VI. 

TIMBER  BEAMS. 

101.  The  Compress! ve   and  Tensile  Strength. 

The  great  number  of  the  different  varieties  of  timber,  and 
the  great  variation  in  the  strength  of  the  timber  that  is 
known  in  different  parts  of  the  globe  by  the  same  name, 
render  the  determination  of  constants  or  mean  values  for  C 
and  T  for  use  in  the  computation  of  the  strength  of  beams  a 
very  difficult  matter,  and  we  find  that  much  difference  exists 
between  the  constants  that  are  given  for  the  same  timber  by 
the  older  and  more  recent  experimenters  ;  especially  is  this  the 
case  for  the  crushing  strength.  The  former  make  C  ~  T 
less  than  0.75  for  all  of  the  principal  varieties,  while  the  latter 
make  it  greater  than  %inity. 

The  computed  transverse  strength  of  the  beams  broken  by 
the  older  experimenters  from  their  constan Ogives,  practically, 
accurate  results,  or  such  as  are  within  the  limits  of  the  varia- 
tion in  strength  of  the  material,  while  those  from  the  more 
recent  experimenters  make  their  computed  transverse  strength 
from  twenty-five  to  one  hundred  per  cent  greater  than  their 
experiments  gave,  which  indicates  that  with  improved  testing 
machines  the  more  recent  experimenters,  such  as  Professor 
Thurston,  Laslet  and  Hatfield,  obtained  their  crushing  strength 
at  a  point  nearer  the  total  destruction  of  the  wood  than  the 
older  with  machines  having  less  power. 

102.  To  Compute  the    Compressive   and  Ten- 
sile Strength.       Crushing. — From  the    fibrous   character 
of  timber  and  its  weak  lateral  adhesion    it  is  difficult  to  de- 


TIMBER   BEAMS.  107 

termine  its  crushing  strength  from  direct  pressure  on  small 
specimens,  or  that  intensity  of  crushing  strain  that  holds  its 
tensile  strength  in  equilibrium  :  this  can  be  accurately  com- 
puted from  Eqs.  31  and  79,  when  the  breaking,  transverse 
and  tensile  strength  have  been  determined  from  experiments. 

EXAMPLE  35.  —  Required  the  Crushing  Strength  of  Teak- 
wood  from  the  known  tensile  strength  and  the  centre  trans- 
verse breaking  load  of  a  rectangular  beam  when 

Depth.  .  .d=2.0  ins.,  T=  15000  Ibs.  mean  of  Mr.  Barlow's  tests, 
Breadth  .5=  2.0  "  Z=938  "  «  "  « 

Span  .  ...  «=Y.O  feet,  m=4. 

The  position  of  the  neutral  line  becomes  from  Eq.  30, 


d  =  8X  2  -  Ag  x  4  x  2  <2)2  1500°  ~  12X8  X  12  X7  =  035  ins 
2  Y  '  4  X  4  X  2  X  150QO 

.  •  .  dc  =  2.0  -  0.35  =  1.65  inches, 
and  the  crushing  value  of  C  from  Eq.  31, 


From  Mr.  Hodgkinson's  experiments  C  =  12100  pounds 
for  Teak. 

Tensile  Strength.  —  This  may  be  computed  from  Eqs.  32 
and  33,  and  from  Eqs.  80  and  81,  when  the  transverse  and 
crushing  strength  have  been  ascertained  from  experiments. 

RECTANGULAR  WOODEN  BEAMS. 

1O3.  Neutral  Line.  Table  of  positions  of  the  neutral 
line  in  rectangular  sections  of  wood,  and  the  mean  ultimate 
crushing  and  tensile  strength  of  the  different  varieties  of 
timber. 


108 


STRENGTH   OF   BEAMS   AND   COLUMNS. 


BREAKING  STRENGTH 
PER  SQUARE  INCH 
IN  POUNDS. 

Ratio  of 
Crushing 
to 
Tenacity, 

?-* 

Depth  of 
Neutral 
Line  Be- 
low the 
Crushed 
Side  of 
the  Beam, 
orrfc. 

Authorities. 

Crushing, 

Tensile, 
T. 

Ash  American 

4400 
5SOO 
8600  H 
5800 
6900 
7700  H 
9300  H 

7000 
4580  H 
6400  H 
6000 
6830 
6500  M 

5850  M 
6000 
7200 
9100 
6500  H 
9500  H 
8000 
8000 
5000 

11000 
14000     ' 
12000  M 
15000 
18000 
90007? 
12000  R 

11600 
11700  R 
15000  M 
10300 
140003/ 
10000  M 

12000  M 
10000 
18000 
18000 
10000  M 
19000  M 
12600 
19200 
1COOO 

0.4 
0.4142 
0.7166 
0.3866 
0.3833 
0.4142 
0.7750 

0.6034 

0.3872 
0.4266 
0.5825 
0.4878 
0.6500 

0.4875 
0.6 
0.4 
0.5055 
0.65 
0.5 
0.6349 
0.4166 
0.5 

0.8900  d 
0.8877  d 
0.826TM 
0.8933d 
0.8941  d 
0.8878  d 
0.8165  d 

0.8477  d 
0.8939  d 
0.8843  d 
0.8518  d 
0.8706  d 
0.8376  d 

0.8711  d 
0.8495  d 
0  8900  d 
0.8673  d 
0.8376  d 
0.8685  d 
0.8410  d 
0.8525  d 
0.8685  d 

B.,  Barlow. 
H.,  Hodgkinson. 
M.,  Moles  worth. 
R.,  Rankine. 

'  t              •« 

"     English 

Beech,  American  .  .  . 

«              n 

"      English 

Birch,         American 
Black  

Birch  English 

«            >< 

Cedar,  American  Red 
Elm         ... 

Fir,  White  Spruce.. 
"         "    Christiana 
Deal.... 
Oak,  American  Red 
"     White 

"    English  

Pine,  Am.  South-) 
ern  Long-leaf.  .  .  \ 
Pine,  Am.  White.  .  . 
Am.  Yellow  .  . 

Red,  European 

"    Dantzic... 
"    Riga  ... 

Poplar  

5400  M 
7500  M 
5400  M 
5748  H 
6586  H 
5100  T 

12000  M 
14000  M 
8000  M 
11549  B 
12857  B 
7000  T 

0.45 
0.5357 
0.575 
0.4977 
0.5122 
0.73 

0.8792  d 
0.8559  d 
0.8533  d 
0.8690  d 
,0.8660  d 
0.8243  d 

Walnut  Black 

. 

The  position  of  the  neutral  lines  in  the  above  Table  was 
computed  from  Eq.  25,  d  being  unity. 

1O4.  Transverse  Strength.  Tim- 
ber beams  breaking  with  a  well-defined 
fracture  in  its  fibres,  the  transverse 
strength  may  therefore  be  computed  from 
the  crushing  and  tensile  strength  by 


TIMBER   BEAMS.  109 

means  of  either  Eq.  27  or  28,  as  the  rectangular  is  the  form 
that  is  principally  used  in  wooden  beams. 

EXAMPLE  36. — Required  the  uniformly  distributed  breaking 
load  of  an  American  White  Pine  Rectangular  Beam,  when 

The  depth d  —  14".0,  C  —     5000  pounds  mean  of  tests, 

"    breadth.... I  =    6".0,  T  =  10000       " 

"    span s  =  28'.  0,  m  =  8,  Art.  34,  and  q  =  0.5. 

The  position  of  the  neutral  line  from  the  Table  is 

dc  —  0.8685  X  14  —  12.16  inches, 
and  the  breaking  load  from  Eq.  27  becomes 

8X6  (12.16)'  5000 
L-         3X12X28         = 

The  Table  in  Trautwine's  Engineers'  Pocket-Book  gives 
37800  pounds  as  the  breaking  load  of  this  beam. 

EXAMPLE  37. — Required  the  breaking  load  of  a  Rectangular 
English  Oak  Beam  fixed  at  one  end  and  loaded  at  the  other, 
when 

The  depth d  —  2".0,  C  =     6500  pounds  mean  by  test, 

"    breadth b  =  2".0,  T  =  10000       "  " 

"    span *  =  4'.0,    q  —  0.65  and  m  =  1,  Art.  34. 

The  position  of  the  neutral  line  is,  from  the  Table, 

</c  =  0.8376  X  2  =  1.6752  inches, 
the  load,  Z,  from  Eq.  27,  becomes 

1X2  (1.6752)'  6500 

L  =  -~~r-^— r^ — —* =  253  pounds. 

3  X  12  X  4 


110  STRENGTH   OF   BEAMS   AND   COLUMNS. 

From  Colonel  Beaufoy's  experiments  the  mean  breaking 
load  of  6  of  these  beams  was  258  pounds  each.* 

EXAMPLE  38. — Required  the  centre  breaking  load  of  the 
beam  described  in  Example  37,  when 

The  span  =  7.0  feet,  m  —  4. 
The  required  load,  Z,  from  Eq.  27,  becomes 

T       4  X  2(1.6752)2  6500 

L  —  —± — i £ =  578  pounds. 

3X  12  X  7 

From  Mr.  Barlow's  experiments  the  mean  breaking  load  of 
three  of  these  beams  was  637  pounds  each. 

EXAMPLE  39. — Required  the  centre  breaking  loads  of  the 
following  Rectangular  Wooden  Beams,  when 

The  depth d  —    2".0,   C  =  values  from  the  Table. 

"    breadth b=    2".Q,T  =       "         "       « 

"    span s  =  50".0,  m  =  4. 

The  position  of  the  neutral  line  is  obtained  by  multiplying 
the  depth,  d,  by  the  factor  corresponding  to  the  ratio  C  -=-  77 
given  in  the  Table,  for  each  case. 

Computed.  2  tests. 

Christiana  Deal.  .L  =  i*lgi?6g>'  585°  -.    969.. .       .  940  and  1052  Ibs. 
o  X  OU 


English  Ash L=         — ax  30  =  1254 1304  and  1304  " 

4X2  (1.7686)2  6400 

English  Birch.  ..L=  —  =  1067 1164  and  1304  " 

3  X  50 

Am.  Black  Birch  L  =   —1073 1027  and  1433  *' 

3  X  50 

These  experimental  breaking  loads  are  taken  from  Mr.  P. 
*  Barlow's  "Strength  of  Materials,"  p.  58. 


TIMBER   BEAMS.  Ill 

W.  Barlow's  experiments  ;*  the  values  for   C  and   T  are  the 
mean  that  are  usually  quoted  by  authors. 

EXAMPLE  40. — Required  the  centre  breaking  load  of  a  Rect- 
angular French  Oak  Beam,  when 

The  depth d=    7.5  inches,  C  =  6000  pounds, 

"breadth I  =    7.5      "       ^=10000     u 

"    span , s  =  15.0  feet,       q  =  0.6  and  m  =  4. 

The  position  of  the  neutral  line  from  the  Table  becomes 

dt  =  0.8495  X  7.5  ==  6.374  inches, 
and  the  required  load,  Z,  from  Eq.  27,  becomes 


L  =  .  =  13M3         ndg 

3X12X15 


M.  Buffon,  in  experiments  made  for  the  French  Govern- 
ment^ broke  two  of  these  beams  with  13828  and  14634 
pounds  respectively ;  from  Rondelet's  experiments  the  values 
of  67  and  T7  are  supposed  to  be  equal  to  those  given  in  the 
Table  for  American  Red  Oak. 

CIRCULAR  WOODEN  BEAMS. 

1O5.  Neutral  Line.  The  position  of  the  neutral  line 
and  the  value  of  the  factors  fc  andyT  for  use  in  Eqs.  76  and 
77  may  be  obtained  from  the  following  Table,  by  proportion 
when  necessary  : 


*  Barlow's  "  Strength  of  Materials,"  p.  86.  f  Ibid.,  p.  56. 


112 


STRENGTH   OF   BEAMS   AND   COLUMNS. 


Ratio  of  Crushing 
to  Tenacity, 
*              or 
C+  T—  q. 

Depth  of  Neutral  Line 
Below  the  Crushed  Side  of 
the  Beam,  or 
dc. 

FACTORS  FOB  COMPUTING  THE  MOMENT 
OF  RESISTANCE,  RADIUS  =  1. 

fc  for  Crushing 
'     Strain,  C. 

/T  for  Tensile 
Strain,  T. 

1.0 

0  6976  d 

1.1548 

1.1548 

0.9829 

0.7000  d 

1.1632 

1.1433 

0.9473 

0.7050  d 

1.1812 

1.1189 

0.9126 

0.7100  d 

1.1991 

1.0943 

0.8790 

0.7150  d 

1.2174 

1.0708 

0.8463 

0.7200  d 

1.2357 

1.0458 

0.8133 

0.7250  d 

1.2540 

1.0200 

0.7837 

0.7300  d 

1.2726 

0.9974 

0.7541 

0.7350  d 

1.2912 

0.9732 

0.7245 

0.7400  d 

1.3098 

0.9490 

0.6965 

0.7450  d 

1.3287 

0.9251 

0.6686 

0.7500  d 

1.3476 

0.9013 

0.6423 

0.7550  d 

1.3666 

0.8775 

0.6161 

0.7600  d 

1.3856 

0.8538 

0.5912 

0.7650  d 

1.4048 

0.8304 

0.5664 

0.7700  d 

1.4240 

0.8066 

0.5429 

0.7750d 

1.4434 

0.7832 

0.5194 

0.7800  d 

1.4628  - 

0.7598 

0.4971 

0.7850d 

1.4828 

0.7365 

0.4749 

0.7900  d 

1.5020 

0.7133 

0.4534 

0.7950d 

1.5216 

0.6895 

0.4320 

0.8000  d 

1.5412 

0.6658 

0.4129 

0.8050  d 

1.5611 

0.6442 

0.3938 

0.8100  d 

1.5810 

0.6226 

1O6.  Transverse  Strength.  The  transverse  strength 
of  circular  wooden  beams  may  be  com- 
puted from  either  Eq.  74  or  75,  also  from 
Eq.  76  or  77,  with  the  aid  of  the  values  of 
the  factors  fc  and  f^  deduced  from  the 
above  Table. 

EXAMPLE   41. — Kequired     the    centre 
breaking  load    of   a  Circular  Christiana 
Deal  Beam,  when 

The  diameter  d  =    2".0,   C  =    5850  pounds  mean  of  tests, 
"    span   ...   s  =  48".0,  T  =  12000       "  "       "      " 

m  =  4,         q=  0.4875. 


TIMBER   BEAMS.  113 

The  required  load,  Z,  becomes  from  Eq.  77, 

T      4  (I)2  0.726  x  12000       ^ 
L  =  -  -  =  726  pounds. 

Mr.  Barlow  broke  three  of  these  beams  with,  740,  796  and 
780  pounds  respectively,  the  mean  being  772  pounds.* 

1C 7.  Relative  Strength  of  Square  and  Circu- 
lar Timber  Beams.  The  required  relation  will  be  ob- 
tained from  either  Eq.  83,  84  or  85,  as  the  case  may  require. 

.EXAMPLE  42. — Required  the  centre  breaking  load  of  a  Cir- 
cular Christiana  Deal  Beam  from  that  of  the  circumscribed 
square  beam  of  the  same  span  and  material,  when 

Side  of  the  square ...  d  =   2".0,  O=    5850  Ibs.  mean  of  tests, 
Diameter  of  the  circle  d=    2".0,  T=  12000  "         "     "      " 

Span s  =  48".0,  q  —  0.4875, 

/*,  from  the  Table,  page  108,  becomes  0.8711  when  d  =  1, 
and  yc,  from  the  Table,  page  112,  becomes  1.4912  when  r  =  1. 

With  these  values  Eq.  83  becomes 

3/c  _     3  X  1.4912 

8/  -  8   X  (0.8711)' 
hence,  Circle  —  Square  X  0.737. 

Breaking  strength  of  2"  sq.  beam,  Mr.  Barlow's  tests,  1117  Ibs. 

"  "          "  2"circ.  "     1117x0.737 823    « 

Mean  breaking  strength  of  3  beams,  Example  39.  ...    772    " 

Mr.  Barlow  says  that  the  2"  square  and  the  2"  diameter  cir- 
cular beams  were  cut  from  the  same  plank,  "  which  was  a  very 
fine  specimen  of  Christiana  deal."  The  breaking  strength  of 
the  2"  square  and  48"  span  beam,  with  the  above  values  of  T 
and  C,  should  have  been  986  pounds,  and  the  circular  beam 
986  x  0.737  =  726  pounds,  as  in  Example  41. 

*  Barlow's  "  Strength  of  Materials,"  p.  78. 


CHAPTEK  VII. 

STRENGTH  OF  COLUMNS. 

1O8.  General  Conditions  of  Failure  of  Col- 
umns. Euler  and  Tredgold  are  credited  with  the  only  well- 
recognized  attempts  that  have  been  made  to  deduce  rational 
formulas  for  the  strength  of  columns,  but  the  basis  of  each 
of  their  theories  involves  the  assumption  of  the  existence  of 
conditions  that  render  their  application  to  the  actual  pheno- 
mena observed  in  practice  inapplicable  without  the  aid  of 
empirical  factors  determined  from  experiments.' 

The  laws  that  govern  the  breaking  strength  of  pillars  were 
investigated,  experimentally,  by  Mr.  Eaton  Hodgkinson  in 
1840,  but  no  rational  theory  has  ever  been  advanced  that  will 
explain  the  phenomena  observed  by  him  and  subsequent  in- 
vestigators; and  our  constructors  are  to  this  time  using  the 
empirical  rules  of  either  Hodgkinson  or  Gordon,  that  were 
deduced  from  the  former's  experiments,  to  compute  the  re- 
quisite dimensions  of  their  pillars,  or  as  they  may  have  been 
modified  to  conform  to  the  results  obtained  by  subsequent  in- 
vestigators from  new  material  or  that  produced  from  improved 
methods  of  manufacture. 

The  effect  that  will  be  produced  upon  a  given  piece  of 
material,  when  subjected  to  a  strain  in  the  direction  of  its 
length,  depends  entirely  upon  its  deflection.  The  deflection 
varies  with  the  material  and  wTith  the  ratio  of  the  length  to 
the  least  diameter.  It  has  been  found,  experimentally,  accord- 
ing to  Mr.  Tredgold,  "  that  when  a  piece  of  timber  is  com- 
pressed in  the  direction  of  its  length,  it  yields  to  the  force  in 
a  different  manner  according  to  the  proportion  between  its 


STRENGTH    OF    COLUMNS.  115 

length  and  the  area  of  its  cross-section,"  which  is  according 
to  the  amount  it  is  able  to  bend  or  deflect  laterally. 

The  material  of  a  pillar  when  subjected  to  an  applied  load 
in  the  direction  of  its  length,  in  order  to  avoid  the  strain  thus 
brought  upon  its  fibres,  deflects  laterally  and  assumes  a  form 
composed  of  one  or  more  curves,  as  its  ends  may  be  round  or 
flat ;  a  round  end  pillar  assumes  the  one  curve  form  and  the 
flat  end  pillar  takes  a  form  made  up  of  three  or  four  curves. 
The  former  is  represented  in  Fig.  34,  and  will  be  called  the 
Triple  Flexure  form  ;  it  is  that  in  which  the  pillar  oifers  the 
least  resistance  to  breaking  by  an  applied  load. 

When  a  pillar  of  the  Triple  Flexure  form  fails  with  deflec- 
tion, tension  exists  in  its  fibres  at  g  (Fig.  34),  and  compression 
at  m  ;  hence  there  must  be  a  point  in  the  line  gam,  at  which 
there  is  no  strain ;  likewise  there  must  be  a  point  without 
strain  in  the  line  gds  ;  and  compression  existing  in  the  fibres 
at  the  points  0,  f  and  &,  the  entire  side,  kef  bo,  must  be  in 
like  condition.  The  fibre  strains  at  two  sections  of  the  pillar, 
such  as  ac  and  db,  must  therefore  increase  uniformly  in  inten- 
sity from  zero  at  a  and  d,  to  its  greatest  at  c  and  b,  respec- 
tively. The  failure  of  a  few  columns  along  the  lines  ae  and 
db  indicates  that  they  join  the  points  of  rJberse  curvature  of 
each  side  of  the  pillar,  though  a  fracture  beginning  at  c 
may  follow  the  line  of  less  resistance,  cm.  There  is  also  a 
point,  n,  in  the  line  g  f,  at  which  the  strain  is  zero  in  intensity, 
and  a  curved  line,  a  n  d,  connecting  these  zero  points  is  the 
neutral  surface  of  the  column. 

Since  the  column  must  bend  symmetrically  there  is  a  point 
in  each  of  the  lines  ac  and  db  that  is  one  fourth  the  length  of 
the  column  from  each  of  its  ends,  mh  and  so,  respectively, 
thus  making  the  middle  curve  of  the  triple  flexure  form  one 
half  of  the  length  of  the  column.  The  angle  that  the  lines 
ac  and  db  make  with  the  plane  of  the  ends  varies  with  the 
material,  but  in  all  pillars  it  is  supposed  to  approximate  the 
angle  of  45  degrees. 


116 


STRENGTH  OF  BEAMS  AND  COLUMNS. 


However  short  a  pillar  may  be  when  loaded,  it  will  attempt 
to  assume  the  triple  flexure  form.  The  curves  above  and  below 
the  lines  ac  and  db  being  the  first  to  take  their  shape,  thus 
give  rise  to  the  various  phenomena  that  are  observed  when 
short  blocks  of  granular  textured  material,  such  as  cast-iron, 
are  crushed.  When  the  points  c,  f  and  b 
coincide,  the  block  is  sheared  across  at  one 
plane ;  when  a  coincides  with  m  and  d 
with  s  the  block  splits  up  into  four  or 
more  wedges. 

The  strength  of  a  pillar  generally  dimin- 
ishes as  it  becomes  longer  in  proportion  to 
its  least  diameter.  Should  a  series  of  pil- 
lars of  an  uniform  section  be  constructed 
with  progressively  increasing  lengths  and 
the  strength  of  each  be  obtained  by  ex- 
periment, there  would  be  found  one  length 
whose  strength  per  square  inch  of  section 
would  be  greatest ;  the  strain  must  be 
uniformly  distributed  over  the  section  of 
the  pillar  and  is  the  crushing  value  of  C 
for  the  material.  As  the  length  increases 
the  strength  decreases,  until  a  length  of 
pillar  is  reached  whose  strength  is  just 
one  half  of  the  greatest  strength  above 
described  ;  the  strain  must  be  uniformly 
varying  in  intensity — zero  at  g  and  great- 
3  4  est  at  y,  where  its  intensity  per  square 

inch  is  the  crushing  value  of  the  material.  The  pillar  of  our 
series,  with  this  ratio  of  length  to  its  least  diameter,  will  fail 
by  crushing  without  deflection  when  it  is  on  the  verge  of  fail- 
ing by  crushing  with  deflection  ;  twice  the  mean  intensity  of 
its  strength  per  square  inch  of  section  is  the  crushing  strength 
of  the  material,  or  that  crushing  strength  that  equilibrates 
the  tenacity  when  beams  and  columns  fail  by  cross-breaking. 


STRENGTH   OF    COLUMNS.  117 

At  the  inception  of  the  deflection  of  the  pillars  of  our  series 
that  fail  with  deflection,  the  neutral  line  coincides  with  the 
side,  mgs,  of  the  pillar ;  as  its  length  continues  to  increase  it 
will  now  be  able  to  deflect,  all  compression  having  been  re- 
moved from  one  side,  and  the  neutral  line  will  move  from  y 
toward  /*,  a  distance  equal  to  the  deflection,  until  the  deflec- 
tion, gn,  equals  that  depth  of  extended  area  that  was  found  to 
be  necessary  for  equilibrium  when  a  beam  of  the  same  material 
and  section  is  broken  by  a  transverse  load,  and  the  pillar  will 
fail  with  all  of  the  phenomena  of  true  cross-breaking  in  a 
beam. 

From  the  length  at  which  a  pillar  of  our  series  just  begins 
to  deflect,  to  that  at  which  it  deflects  sufficiently  to  cross-break, 
the  anomaly  of  the  strength  of  pillars,  increasing  with  an  in- 
crease of  length,  is  exhibited  ;  of  two  pillars  of  the  same  section 
and  material,  the  one  that  deflects  most,  if  it  breaks  within 
the  limits  of  deflection  above  described,  will  sustain  the 
greater  load.  When  the  length  is  increased  beyond  that  at 
which  the  pillar  first  breaks  by  true  cross-breaking,  its  strength 
again  diminishes,  and  the  further  anomaly  is  presented  of  a 
column  failing  by  cross-breaking  with  the  same  load  that  a 
column  of  the  same  section  and  shorter  length  will  fail  with 
by  crashing,  but  with  less  deflection. 

The  pillar  by  deflecting  seeks  to  relieve  itself  of  the  resting 
of  the  load  directly  on  its  middle  section,  but  this  it  can  only 
do  to  the  extent  of  gn  =•  r/T,  or  that  value  required  for  rupture 
in  beams,  and  it  must  fail  by  the  direct  pressure  of  the  load 
along  the  line  nf,  as  it  is  being  transmitted  to  the  foundation, 
so,  for  all  lengths  of  pillars ;  but  when  it  deflects  so  much  that 
the  resultant  of  the  applied  load  passes  without  the  middle 
section  of  the  pillar,  it  thus  increases  its  effect  by  a  cantilever 
strain,  which  effect  must  be  deducted  from  the  amount  of  the 
two  pressure  wedges  that  are  required  along  nf  to  rupture  the 
fibres  by  both  tension  and  compression,  as  explained  in  Art. 
109. 


118  STRENGTH    OF   BEAMS    AND    COLUMNS. 

PIN  END  COLUMNS. 

The  "  failure "  of  long  pin  end  columns  or  those  that  fail 
by  cross-breaking^  presents  some  interesting  peculiarities.  The 
frictional  resistance  offered  to  the  movement  of  the  pillar 
around  the  pin,  as  it  deflects,  increases  with  the  size  of  the 
pin  ;  it  varies  in  pillars  of  like  dimensions  and  material  when 
tested  with  pins  of  the  same  size,  and  decreases,  in  all  cases, 
after  the  pillar  has  been  fairly  set  in  motion  by  the  lateral  de- 
flection. Thus,  the  results  of  experiments  show  that  a  load 
less  than  that  required  to  produce  fracture  by  compression 
will  be  sustained,  but  as  the  deflection  approaches  near  to  dT 
in  value,  the  amount  of  friction  becomes  so  small  that  the  pil- 
lar will  "  suddenly  spring  "  to  a  deflection  that  will  cause  the 
pillar  to  fail  with  this  load  by  true  cross-breaking,  and  of  two 
like  columns,  one  will  sustain  for  a  time  a  greater  load  than 
the  other,  on  account  of  greater  frictional  resistance  around 
the  soffit  of  the  pin,  evidenced  by  its  deflection  being  the  least ; 
the  greater  load  will  then  commence  to  decrease,  with  a  freer 
motion  of  the  pillar,  and  will  finally  reach  a  point  where  it 
will  "  suddenly  spring "  to  a  cross-breaking  deflection,  thus 
breaking  with  a  less  load  than  it  had  already  sustained,  but  with 
less  deflection.  That  the  pillar  is  truly  broken  by  its  "  sudden 
spring  "  is  evidenced  by  the  fact  that  upon  being  released  from 
the  pressure  of  the  load  it  will  now  require  with  wrought-iron 
pillars,  in  many  cases,  less  than  one  half  of  the  original  load 
to  produce  the  greatest  deflection  before  obtained.  This  un- 
certain and  variable  frictional  resistance  around  the  pins  of 
pin  end  columns  gives  rise  to  many  anomalous  results  that 
would  not  otherwise  be  encountered  in  making  a  series  of 
tests  with  pin  end  pillars  when  made  of  homogeneous  material, 
such  as  wrought-iron,  and  renders  it  impossible  to  determine 
the  exact  law  that 'governs  their  failure  without  eliminating  its 
irregular  effect.  The  momentum  of  the  sudden  spring  fre- 
quently causes  the  observed  deflection  to  greatly  exceed  the 


STRENGTH    OF   COLUMNS.  119 

true  deflection  at  failure,  and  the  load  sustained  at  the  instant 
of  springing  often  exceeds  the  true  cross-breaking  load.  These 
sources  of  erroneous  observation  become  less  in  effect  as  the 
size  of  the  pin  is  increased  and  as  the  pillar  becomes  shorter 
m  length,  and  to  increase  in  effect  with  an  increase  in  the 
length  of  the  pillar  and  a  decrease  in  the  size  of  the  pin. 

1O9,  Resistance  of  the  Cross-Section  of  the 
Column.  From  the  preceding  Art.  it  is  apparent  that  when 
a  pillar  fails  by  crushing  without  deflection,  its  strength  is 
simply  the  product  of  its  area  by  the  mean  intensity  of  the 
pressure  per  square  inch,  which  varies  with  its  length,  and  that 
after  it  begins  to  deflect,  equilibrium  must  be  established  be- 
tween the  moments  of  the  applied  and  resisting  forces  with 
reference  to  the  fulcrum  or  origin  of  moments^  (Fig.  34),  the 
neutral  line  being  at  n,  the  line  of  direction  of  the  forces, 
perpendicular  to  the  section  fg,  and  uniformly  varying  in  in- 
tensity. 

When  a  column  fails  with  a  deflection  that  compels  the  re- 
sultant of  the  applied  load  to  pass  through  the  middle  section  of 
the  pillar,  it  will  only  be  necessary  to  find  the  amount  of  direct 
pressure  along  the  line  ft/ that  will  cause  the  pillar  to  "  fail."  The 
tensile  strain  along  the  line  gn  must  be  held  in  equilibrium  by 
an  uniformly  varying  crushing  strain  along  the  line  nf\  the 
maximum  intensity  of  this  pressure  wedge  is  always  less  than 
the  crushing  value  of  C  for  the  material,  until  gn  =  d^  which 
is  required  for  rupture  in  a  beam,  when  it  becomes  equal  to 
it.  It  will  also  require  another  pressure  wedge  along  this 
same  line  nf  to  crush  the  material,  independently  of  the  pres- 
sure wedge  that  is  held  in  equilibrium  by  the  tensile  strain 
along  the  line  gn  ;  hence  the  actual  pressure  required  along  the 
line  nf  is  the  sum  of  the  two  pressure  wedges,  one  that  crushes 
the  material  direct,  and  tne  other  that  ruptures  the  extended 
portion  of  the  pillar  ng.  "When  gn  =  d^  these  two  pressure 
wedges  become  equal  in  size,  and  they  must  be  constant  in  value 


120  STRENGTH    OF   BEAMS   AND    COLUMNS. 

for  all  lengths  of  pillars  that  fail  by  true  cross-breaking,  and 
their  sum  is  the  breaking  load  of  the  pillar  as  long  as  the  re- 
sultant of  the  applied  load  passes  through  the  middle  section 
of  the  pillar. 

The  uniformly  distributed  load  on  the  top  of  the  pillar  at 
m&,  by  the  deflection,  has  been  converted  into  an  'uniformly 
varying  load  at  the  section  fg ;  it  will  so  continue  when  the 
pillar  becomes  a  cantilever,  being  zero  in  intensity  next  the 
concave  side  of  the  pillar  and  greatest  at  the  tangent  line  con- 
necting the  points  Jc  and  #,  and  its  moment  or  power  to  break 
the  pillar  as  a  cantilever  must  be  computed  with  reference  to 
the  fulcrum  /",  its  lever-arm  will  be  (tf  -- -  \d)  for  the  rectangle, 
(d  —  \d)  for  the  solid  circle,  and  (8  —  J<#)  for  hollow  circles, 
in  which  d  is  the  least  diameter  and  d  the  deflection.  The 
direct  tensile  and  compressive  strain  along  the  line  gf  that 
is  required  to  hold  in  equilibrium  the  moment  of  the  applied 
load  acting  upon  the  pillar  as  a  cantilever  must  be  deducted 
from  the  sum  of  the  two  pressure  wedges  that  are  required 
to  cross-break  the  pillar,  in  order  to  obtain  its  true  breaking 
load. 

Columns  can  be  constructed  of  wrought-iron,  steel,  wood, 
and  some  cast-iron  that  will  "  fail "  by  cross-breaking,  with 
the  full  value  of  the  sum  of  these  two  equal  pressure  wedges  ; 
but  with  most  cast-iron  and  other  material  in  which  C  -r-  T  is 
approximately  greater  than  1.75  for  solid  circles,  3.00  for  hol- 
low circles  and  2.0  for  rectangles,  the  pillar  becomes  a  canti- 
lever before  it  deflects  sufficiently  to  cross-break. 

The  sum  of  these  equal  pressure  wedges  is  the  greatest  load 
that  a  pillar  can  bear  at  the  instant  of  "  failure  by  cross-break- 
ing," but  experimenters  frequently  obtain  for  short  pillars  a 
greater  apparent  breaking  load,  from  the  fact  that  as  the  re- 
sultant of  the  load  passes  through  its  middle  section,  the  pillar 
cannot  at  the  instant  of  tow  failure  escape  from  under  it,  and 
new  conditions  are  set  up,  such  as  making  two  pillars,  etc. 

The  formulas  deduced  for  failure  of  pillars  by  cross-break' 


STRENGTH    OF   COLUMNS. 

ing  from  this  theory  are  identical  in  form  with  those  given  by 
Mr.  Lewis  Gordon,  and  it  can  thus  be  seen  why,  that  although 
his  formulas  are  deduced  from  impossible  conditions  of  cross- 
breaking,  they  can  be  so  modified  by  the  teachings  of  experi- 
ment that  very  accurate  results  may  be  obtained  from  their 
use. 

110,  Notation.       The    following     notation    has    been 
adopted   from   previous   chapters,    and   will   have   the    same 
significance  wherever  it  is  used.     Other  special  notation  will 
be  given  as  it  may  be  required. 

C  =  the  greatest  intensity  of  the  compressive  strain  in  Ibs.  per  square  inch. 
T  =  the        "  "      tensile  "         "       "         "          " 

q  =.  the  quotient  arising  from  dividing  C  by  T. 

b  =  the  width  of  the  section  in  inches. 

d  —  the  depth  in  the  direction  in  which  flexure  takes  place  in  inches. 
dc  —  the  depth  of  the  compressed  area  in  inches, 
dr  =  the      "         "      tension 
L  =  the  breaking  load  in  pounds. 

I  —  the  length  of  the  pillar  in  inches. 

d  =  the  deflection  in  inches. 

111.  Deflection  of  Columns.     In  the  present  state 
of  our  knowledge  of  the  cause  of  flexure,  the  laws  that  deter- 
mine the  amount  that  a  pillar  will  deflect  at  the  instant  of 
failure  can  only  be  obtained  from  a  carefully  conducted  series 
of  experiments.     On  account  of  the  expense   and  the  great 
pressure  required,  the  number  of  experiments  on  the  breaking 
strength  of  pillars  of  the  size  used  in  structures  is  very  small 
and  incomplete  for  all  lengths  required  for  a  complete  knowl- 
edge of  the  subject ;  and  as  the  theorists  have  taught,  following 
Euler,  that  the   strength  of  a  column  is  independent  of  its 
deflection,  it  has  caused  the  experimenters  to  so  regard  it  and 
pay  it  but  little  attention,  their  means  of  measuring  it  being 
crude,  and  the  record  made  in  an  unsatisfactory  manner. 

The  amount  of  deflection  that  is  required,  before  a  column 


122  STRENGTH   OF   BEAMS   AND   COLUMNS. 

can  be  broken,  is,  comparatively,  a  very  small  quantity,  much 
less  than  that  required  for  rupture  should  the  pillar  be  broken 
as  a  beam,  and  requires  very  accurate  means  for  its  measure- 
ment. The  amount  that  a  pillar  deflects  is  usually  stated  to 
be  the  distance  that  its  middle  section  "  moves,"  measured 
from  a  given  stationary  point,  which  is  only  correct  when  the 
points^,/*  and  o  (Fig.  34)  lie  in  the  same  right  line  at  the 
beginning  of  the  movement ;  should  f  lie  to  the  right  of  the 
line  ko,  the  observed  deflection  will  be  too  great  by  its  distance 
from  the  line,  and  when  /'  lies  to  the  left  of  the  same  line  #<?, 
the  observed  deflection  will  be  too  small  by  its  distance  from 
the  line  £<?,  which  must  be  parallel  to  the  line  of  direction  of 
the  resultant  of  the  applied  load.  The  deflection  is  an  inci- 
dental quantity  ;  the  distance  required  is  the  lever-arm  of  the 
resultant  of  the  load,  which  is  the  perpendicular  distance  from 
the  fulcrum,  y,  to  its  line  of  direction. 

I    =  the  length  of  the  pillar  in  feet  and  decimals, 
dc  =  the  depth  of  the  compressed  area  in  inches, 
c    =  a  constant  quantity  determined  from  experiments. 
#  =  the  breaking  deflection  in  inches. 

Then 

$  =  cjr.  (93) 

ac 

The  factor,  <?,  must  be  determined  for  the  different  material 
used  in  structures,  from  experiments  on  fixed,  round,  two  pin, 
and  on  one  pin  and  one  fixed  end  columns  ;  it  has  a  different 
value  for  each  material  and  style  of  end  connections,  and 
probably  for  each  different  shape  of  cross-section. 

EXPERIMENTS. 

The  experiments  required  to  determine  the  laws  that 
govern  the  deflection  of  columns  of  a  given  section  and 
material  may  be  very  much  simplified  by  determining  defi- 
nitely the  following  points : 


STRENGTH    OF   COLUMNS.  123 

1st.  The  ratio  of  I  -4-  d  that  gives  the  true  crushing  value 
of  the  material. 

2d.  The  largest  ratio  of  /  -f-  d  that  sustains  its  greatest 
load  without  deflection. 

3d.  The  smallest  ratio  of  I  -r-  d  that  cross-breaks. 

4th.  The  largest  ratio  of  I  -=-  <$  that  cross-breaks  without 
leverage  of  the  applied  load. 

The  law  governing  the  change  in  the  strength  from  the  1st 
to  the  2d  and  from  the  2d  to  the  3d  ratio  would  then  be  re- 
quired ;  from  the  3d  to  the  4th  ratio  the  strength  would  be 
constant,  and  for  larger  ratios  than  the  4th  the  deflection,  or 
lever-arm  of  the  load,  would  be  the  onlj  varying  element,  and 
only  for  these  pillars  will  it  be  necessary  to  determine  the 
values  of  the  factor  c,  in  Eq.  93,  for  the  various  kinds  of 
material  used  in  structures. 

JSTo  attempt  will  be  made  to  determine,  for  any  given 
material,  the  limits  above  defined,  from  the  incomplete  record 
of  experiments  at  present  existing. 

112.  Classification  of  Pillars.  From  the  descrip- 
tion of  the  manner  of  "  failure "  of  columns  of  various 
lengths,  it  is  evident  that  they  may  be  divided  into  the  follow- 
ing general  classes : 

1st.    PlLLARS    THAT    FAIL    BY    CRUSHING. 

2d.    PlLLAKS    THAT    FAIL    BY    CROSS-BREAKING. 

These  classes  may  be  again  divided  into  Cases,  which,  for 
convenience  of  reference,  will  be  numbered  consecutively. 

Pillars  that  fail  by  crushing. 
Case  I. — PILLARS  THAT    FAIL  WITH    THE    FULL    CRUSHING 

STRENGTH    OF    THE    MATERIAL. 

Case  II. — PILLARS  THAT  FAIL  WITH  LESS  THAN  THE  FULL 

CRUSHING    STRENGTH    OF    THE    MATERIAL    AND    WITHOUT     DEFLEC- 
TION. 


124  STRENGTH   OF   BEAMS   AND    COLUMNS. 

Case  III. — PILLARS    THAT    FAIL    WITH    LESS   THAN    THE 

FULL    CRUSHING    STRENGTH     OF     THE     MATERIAL     AND    WITH     DE- 
FLECTION. 

Pillars  that  fail  ~by  cross-breaking. 
Case  IV» — PILLARS  THAT  CROSS-BREAK  FROM  COMPRESSION. 

Case  V. — PILLARS  THAT  CROSS-BREAK  FROM  COMPRESSION 
AND  CANTILEVERAGE. 

END  CONNECTIONS. 

Columns  are  again  classed  from  the  form  or  shape  of  their 
end  connections,  while  they  all  fall  under  one  or  the  other  of 
the  five  cases  given  above.  In  columns  of  a  given  material  and 
dimensions,  the  only  element  of  variation  in  their  strength  aris- 
ing from  differences  in  its  end  connections  is  the  deflection. 

A  flat  or  fixed  end  column  is  one  whose  ends  are  planes 
at  right  angles  to  its  length  ;  for  a  given  material  and  dimen- 
sions it  deflects  less,  and  is,  therefore,  the  form  of  greatest 
strength. 

A.  round  end  column  is  one  whose  ends  are  spherical ;  con- 
sequently, being  free  to  move  laterally,  it  deflects  most  and  is 
the  form  of  least  strength. 

Pin  end  is  a  class  of  columns  used  in  bridge  construction 
in  which  the  load  is  applied  to  pins  passing  through  holes  in 
its  head  and  foot,  and  by  them  transmitted  to  the  column. 
The  smaller  the  pin  and  the  less  frictional  resistance  that  may 
be  developed  between  the  pin  and  its  soffit,  the  nearer  it  ap- 
proaches the  condition  of  a  true  round  end  column,  and  the 
less  strength  it  will  exhibit,  while  the  larger  the  pin  and  the 
greater  frictional  resistance  that  may  be  developed  the  nearer 
it  approaches  the  condition  of  a^atfend  pillar,  and  the  greater 
strength  it  will  exhibit. 

One  pin  and  one  flat  end,  in  its  strength,  follows  laws 
similar  to  the  above  and  a  mean  between  flat  and  round  end 
pillars. 


STRENGTH    OF   COLUMNS.  125 

MATERIAL  AND  SECTION. 

In  practice,  columns  are  again  classed  as  wooden,  cast-iron, 
wrought-iron  and  steel,  from  the  material  of  which  they  are 
made,  and  from  the  shape  of  their  cross-sections  into  rect- 
angular, circular,  hollow  circular,  angle-iron,  box,  channel,  eye- 
beam,  tee,  etc.  From  the  incomplete  tests  at  present  made 
the  lengths  of  columns  of  a  given  section  and  material  that 
belong  to  each  of  the  five  cases  of  failure  given  above  cannot 
be  determined.  In  the  examples  quoted  in  the  sequel  these 
limits  are,  however,  well  defined  in  some  special  cases  of 
material. 

113.  Case  I. — COLUMNS  THAT  FAIL  WITH  THE  FULL  CRUSH- 
ING  STRENGTH  OF  THE  MATERIAL. 

Let  A  represent  the  area  of  the  section  of  the  pillar  in 
square  inches,  then  as  the  load  must  be  uniformly  distributed 
for  this  case,  the  crushing  load  will  be 

L  =  A  X  C.  (94) 

114.  Case  II. — COLUMNS  THAT  FAIL  WITH  LESS  THAN  THE 

FULL    CRUSHING  STRENGTH    OF    THE    MATERIAL   AND    WITHOUT    DE- 
FLECTION. 

In  all  pillars  of  this  class,  at  failure,  the  strain  at  the  surf  ace 
of  one  side  is  constant  and  equal  to  the  crushing  value  of  C 
for  the  material ;  the  strain  at  the  surface  of  the  opposite  side 
varies  with  the  length,  from  the  crushing  value  of  C  to  zero  in 
intensity,  at  which  length  the  strain  is  uniformly  varying,  the 
mean  intensity  being  C  -=-  2. 

1Q  =  the  length  of  pillar  that  fails  with  full  crushing  strength, 
/  =  the       "       "       "        "      "        "  a  mean  intensity,  <7-h  2, 
x  —  the       "       "       "      whose  strength  is  required, 
A  =  the  area  of  the  section  of  the  pillar  in  square  inches. 

The  values  of  Z0  and  I  vary  with  the  material  and  are  de- 
termined experimentally  ;  assuming  that  the  intensity  of  the 


126  STRENGTH   OF   BEAMS   AND   COLUMNS. 

strain  on  one  side  of  the  pillar  varies  with  the  length,  then  for 

(^ /2>\ 
J  and  the  mean  for  the  sec- 
tion will  be  f  C -\-  Gij v)  J  -r-   2,  and  the  load  that  the 

pillar  of  the  length  x  will  fail  with  is 


For  the  longest  column  to  which  this  formula  applies,  .1  —  X, 
and  Eq.  95  becomes 

L  =  ^-*L-  (95A) 

To  Compute  the  Crushing  Strength.  —  The  above  described 
manner  of  failure  of  pillars  and  the  formulas  resulting  may 
be  used  very  advantageously  to  compute  the  crushing  strength 
of  any  material,  which,  in  this  case,  is  not  affected  by  the 
tensile  strength.  And  we  are  thus  furnished  with  a  valuable 
means  of  testing  the  correctness  of  the  computed  crushing 
strength  for  the  same  material  when  broken  in  a  beam  by  the 
methods  given  in  Articles  39  and  55  ;  also  that  obtained  from 
direct  experiments  with  short  blocks. 

Deducing  the  value  of  C  from  Eq.  9  5  A,  we  have 

C=2-,  (95B) 


which  is  the  formula  required  from  which  to  compute  the 
crushing  strength  of  the  material  in  a  pillar  that  fails  or  sus- 
tains its  greatest  load  without  deflection  when  it  is  on  the 
verge  of  failing  with  deflection. 

EXAMPLE  43.  —  Required  the  greatest  strength  of  a  Rect- 
angular Wrought-Iron  Column,  tested  with  two  \"\  pin  ends, 
when  the  deflection  d  —  0,  area  A  —  8.85  square  inches,  and 


STEENGTH   OF   COLUMNS.  •  127 

0  —  50000,  assumed  to  be  equal  to  the  tensile  strength  ob- 
tained from  direct  tests. 
From  Eq.  95 A  we  have 

8.85  X  50000 
L  —  -  -  —  221250  pounds. 

2 

From  the  United  States  Government  "Watertown  Arsenal 
Tests,  1882-1883,*  the  greatest  strength  of  nine  of  these  col- 
umns was  found  by  experiments,  the  mean  was  234850  pounds, 
the  deflection  varied  from  0".02  to  0".12,  the  mean  wasO".07; 
the  lengths  varied  from  54  to  78  inches,  and  they  were  all 
approximately  three  inches  square.  The  mean  strength  of 
four  of  these  columns,  tested  with  one  flat  and  one  \"\  pin 
end  was  210800  pounds,  the  mean  deflection,  0".12 ;  the 
length  of  two  was  90  inches  and  that  of  the  other  two  columns, 

O  ' 

120  inches. 

The  mean  strength  of  four  of  these  columns,  tested  with  flat 
ends,  was  217875  pounds,  the  mean  deflection  was  0".13,  the 
lengths  were  90  and  120  inches. 

115.  Case  III. — COLUMNS  THAT  FAIL  WITH  LESS  THAN  THE 

FULL  CRUSHING    STRENGTH  OF    THE    MATERIAL    AND  WITH  DEFLEC- 
TION. 

The  pillars  of  this  class  vary  in  length  for  any  material  and 
section,  between  the  limits  of  I  of  Case  II.,  or  that  length  at 
which  the  mean  intensity  of  the  crushing  load  is  C  -r-  2,  and 
that  length  of  Case  IY.  that  first  fails  by  cross-breaking,  both 
limits  being  determined  by  experiment. 

T1  =  the  greatest  tensile  strain  at  the  middle  section. 

C'  =  the        "        compressive  strain  required  to  balance  T '. 

Then  assuming  that  the  intensity  of  the  tensile  strain  in  the 
convex  side  of  the  pillar  varies  with  the  deflection,  d,  from  0 

*  Senate  Ex.  Doc.  No.  5— 48th  Congress,  1st  Session,  pp.  60-67. 


128  STRENGTH    OF   BEAMS   AND   COLUMNS. 

to  T7,  the  tenacity  of  the  material,  the  following  formulas  are 
deduced  for  the  various  sections  : 


d.f  ' 
6T 


By  giving  to  q  and  d  the  proper  value  for  the  section,  the 
following  formulas  have  been  deduced  : 

RECTANGULAR  PILLARS. 
1         From  Eqs.  27  and  28,  page  88,  we  have 


J  t/T    IZ/CC/    —I—     </P)       .  1     •      1  7  .p. 

"•  '  q  =  -      —~ -,  in  which  «T  =  o. 

di  4 

Then  the  expression  for  O '  becomes,  from 
substituting  these  values, 

tf2  (3^  _  $\  T 


„, 

</T  (d  -  07   ' 

The   sum  of  the  greatest  intensities  of  the  two  pressure 
wedges  at  "  failure"  is  C  +  C'. 


The  value  of  dT  in  this  formula  is  that  required  for  rupture 
of  the  column  when  broken  as  a  beam. 

EXAMPLE  44.  —  Required  the  'greatest  strength  of  a  Rect- 
angular Wrought-Iron  Column,  tested  with  two  \"\  pin 
ends,  when 

The  diameter  .......  d  =  3".  00,   C  —  50000  pounds  assumed, 

"     breadth  ........  I  =  3".  00,  T  =  50000       "       by  test, 

"     deflection  ......  d  =  0".34,  ^  =  0".66  for  rupture. 


STRENGTH    OF    COLUMNS.  129 


From  Eq.  96  we  have 


L  =  [50000  +  ^.soooo  =  «  poimd, 


The  greatest  strength  was  284000  pounds,*  the  length  was 
30  inches. 

CIRCULAR  PILLARS. 

By  substituting  C  +C'  for  2(7  in  Eq.  100,  page  134,  which 
gives  the  resultant  of  the  load  for  this  case,  we  have  for  the 
load  that  the  pillar  will  fail  with, 


L  =  \  C  -M~ 


The  numerical  values  required  for  the 
factors  q  and  f  are  to  be  taken  from  the 
Table,  page  134,  for  that  position  of 
the  neutral  line  that  corresponds  to 
dc  —  d  —  8.  The  value  of  dT  is  that  re- 
quired for  rupture  of  the  material  in  a 
circular  beam,  with  its  full  compressive 
and  tensile  strength. 

EXAMPLE  45. — Required  the  greatest  load  sustained  by  a 
Cylindrical  Column  of  Midvale  Steel,  when 

The  diameter  d  =  1".129,  C  =  152000  pounds  mean  of  tests, 
"     length ...  1  =  8".96,    T7  =  112285       "  "      "      " 

"     deflection  d  =  0".25,     q  =  1.353. 

From  the  Table,  page  134,  /  =  1.1839,  for  the  neutral  line" 
at  failure,  dc  =  1".129'  -  0.25,  q  =  0.5278,  and  dT  =  0.38, 
the  position  of  the  neutral  line  of  rupture. 

*  From  the  United  States  Government  Watertown  Arsenal  Tests,  1888, 
Senate  Ex.  Doc.  No.  5—  48th  Congress,  1st  Session,  page  56. 


130  STRENGTH   OF   BEAMS   AND    COLUMNS. 

From  Eq.  97  we  have 

L  =  (152000  +  025X°-5^X112285)  (0.6645)'  1-1839  =  72056  pounds. 
\  U.oo  / 

The  greatest  strength  was  81250  pounds.* 

HOLLOW  CIRCULAR  PILLARS. 

By  substituting  C+  C'  for  2<7  in  Eq.  101,  page  135,  which 
gives  the  resultant  of  the  load  for  this  case,  we  have 

,,  (98) 

in  which  the  numerical  values  required  for  q  and  f0  are  to  be 
taken  from  the  Table,  page  136,  for  that 
position  of  the  neutral  line  at  failure  that 
corresponds  to  dc  =  d  --  #,  the  radius 
of  the  outer  circle  being  r  and  t  the 
thickness  of  the  metal,  both  to  be  ex- 
pressed in  inches.  The  numerical  value 
of  dT  is  that  required  for  rupture  of  the 
material  in  a  hollow  circular  beam,  with 

its  full  compressive  and  tensile  strength. 

116.  Case  IV.—  COLUMNS  THAT  CROSS-BREAK  FROM  COM- 
PRESSION. 

The  shortest  pillar  of  this  class,  for  a  given  section  and 
material,  is  that  (Fig.  34)  in  which  gn  =  <5  =  dn  or  that 
depth  of  extended  area  required  for  equilibrium  when  a  beam 
of  this  section  and  material  is  broken  by  a  transverse  load,  and 
the  longest  is  that  that  just  deflects  sufficiently  to  allow  the  re- 
sultant of  the  applied  load,  Z,  to  pass  through  \hv  fulcrum,  f ; 
this  varies  with  the  section,  for  a  given  material. 

*  From  the  United  States  Government  Watertown  Arsenal  Tests  for 
1883-84,  Senate  Ex.  Doc.  No.  35— 49th  Congress,  1st  Session,  p.  369. 


STRENGTH   OF   COLUMNS.  131 

The  load  is  the  sum  of  the  two  pressure  wedges.  In  rect- 
angular areas  or  those  that  may  be  divided  into  rectangular 
areas,  the  load  is  the  compressed  area,  multiplied  by  the  crush- 
ing strength  of  the  material. 

RECTANGULAR  PILLARS. 

The  position  of  the  neutral  line  must 
be  computed  from  either  Eq.  25  or  26, 
page  38.  Then,  in  order  that  the  breaking 
load  of  a  given  pillar  may  be  deduced 
from  the  principles  applicable  to  Case  IY., 
tf,  the  deflection,  must  be  equal  to  or 
greater  than  6?T,  and  equal  to  or  less  than 
Id. 

Then 

L  =  bd0C,  (99) 

from  which  the  required  load  may  be  computed,  being  the  sum 
of  the  two  pressure  wedges  required  for  rupture. 

EXAMPLE  46.— Required  the  breaking  load  of  a  Rectangular 
Yellow  Pine  Column,  tested  with  flat  ends,  when 

The  diameter..^  =  5".5,     C  =  5230  pounds  mean  of  four  tests, 
"    breadth.. J^=5".5,     T  =  15478      "  "      "tests, 

"    deflection  tf  =  0".66,   q  ^    0.337. 

The  position  of  the  neutral  line,  computed  from  Eq.  26, 
page  38,  is  dc  =  5". 39,  and  dT  =  0.11,  being  less  than  the  de- 
flection, the  column  failed  by  cross-breaking ;  the  deflection 
being  less  than  one  third  of  the  least  diameter,  it  failed  with- 
out cantileverage. 

.  • .  L  =  5.5  X  5.39  X  5230  =  155043  pounds. 

In  a  series  of  experiments  to  test  the  strength  of  yellow  pine 
columns  with  flat  ends,  conducted  on  the  United  States  Gov- 


132 


STRENGTH  OF  BEAMS  AND  COLUMNS. 


eminent  testing  machine,  at  Watertown  Arsenal,  1881-2,* 
thirty  columns  were  broken,  whose  lengths  varied  from  I  =  %7d 
to  I  =  45d,  which  appears  to  be  the  limits  of  cross-breaking, 
without  leverage  for  the  yellow  pine  tested.  The  crushing 
strength  obtained  from  short  blocks  differed  so  greatly  that  a 
mean  was  not  admissible,  though  two  consistent  classes  can 
be  made  of  the  material  from  these  tests.  The  mean  of  four 
tests  was  C  =  5230  pounds,  from  which  the  required  load  in 
the  above  example  was  computed  ;  the  mean  load  from  three 
tests  was  154000  pounds.  The  mean  value  of  C  from  three 
other  tests  was  3600  pounds,  from  which  the  computed  load 
in  the  following  Table  was  obtained.  The  tensile  strength  was 
obtained  on  the  same  machine  the  previous  year,  and  not 
from  the  material  composing  the  columns  tested. 

The  "  experimental "  load  in  the  Table  is  a  mean  of  the 
number  of  tests  given  in  the  last  column  ;  from  "  failure  at 
knots  and  diagonally,"  the  remainder  of  the  thirty  experiments 
mentioned  above  were  rejected. 


DIMENSIONS  IN  INCHES. 

LOAD  IN  POUNDS. 

/  -t-  (I. 

Deflection  in 

No.  of 

I. 

6. 

d. 

Inches. 

Computed. 

Experiment'  1 

Tests. 

27.0 

180 

15.6 

6.6 

.5  to  .62 

344822 

409000 

3 

30.8 

180 

12.0 

5.8 

.73  "  1.26 

235000 

250000 

1 

81.2 

240 

9.67 

7.7 

.82  "  1.6 

249253 

281000  j   2 

31.2 

180 

15.5 

5.6 

.52 

290718 

344000 

2 

32.8 

180 

5.5 

5.5 

.52 

101574 

129500 

2 

36.0 

180 

12.1 

5.0 

1.3  '•'  1.8 

202554 

230000 

1 

38.2 

210 

5.5 

5.4 

1.03  "  1.95 

100584 

97330 

3 

40.0 

180 

11.6 

4.4 

.9  "  1.30 

170800 

126350 

2 

43.0 

320 

9.28 

7.4 

.99  "  2.55 

231516 

199830 

3 

45.0 

240 

5.4 

5.4 

1.28 

98776 

84900 

2 

45.0 

180 

11.35 

4.1 

.95 

166394 

142000 

1 

In  the  above  Table  the  second  and  sixth  tests  carried  the 
maximum  load  with  deflections  that  varied  a  half  an  inch  in 
amount  before  failure. 


*  Senate  Ex.  Doc.  No.  1— 47th  Congress,  2d  Session,  p.  321. 


STRENGTH   OF   COLUMNS. 


133 


EXAMPLE  47. — Required  the  breaking  loads  of  the  series 
of  White  Pine  Columns  tested  with  flat  ends,  on  the  United 
States  Government  machine,  at  Watertown  Arsenal,  1881-2,* 
when  T  =  10000  pounds  mean  of  other  tests,  C  —  2500  pounds 
mean  of  five  of  these  tests.  The  position  of  the  neutral  line 
from  Eq.  26  for  q  =  0.25  is 

de  =  0.92^. 

The  breaking  loads  in  the  following  Table  were  computed 
from  Eq.  99,  as  in  Example  46  ;  the  first  pillar  of  the  Table  did 
not  deflect  quite  sufficiently  to  belong  to  this  class  of  columns, 
but  as  the  error  is  small,  the  lower  limit  is  taken  to  be,  as  in  yel- 
low pine,  I  =  276? ;  the  greater  limit  was  not  determined  by 
the  tests. 

The  "experimental loads  "  and  the  dimensions  are  the  means 
of  those  given  for  three  tests. 


DIMENSIONS  IN  INCHES. 

LOAD  IN  POUNDS. 

l  +  d. 

I. 

b. 

d. 

Inches. 

Computed. 

Experimental. 

27 

180 

15.6 

6.66 

.4   to  .3 

239400 

227300 

32 

180 

15.6 

5.62 

.37  *   .83 

202800 

164100 

32 

240 

9.4 

7.45 

.7   '  1.2 

161200 

170300 

33 

180 

11.3 

5.4 

.6 

165400 

156200 

36 

280 

9.6 

7.69 

.66  '  1.2 

169900 

153300 

40 

180 

11.6 

4.48 

.43  '  1.25 

119700 

130700 

43 

320 

9.3 

7.47 

.7   '  1.67 

159700 

147600 

CIRCULAR  PILLARS. 

For  this  Case  the  deflection,  #,  must  be  equal  to  or  greater 
than  d^  the  depth  of  the  tension  area  given  by  the  Table  be- 
low for  the  neutral  line,  when  C  -f-  T  =  </,  for  the  material, 
and  equal  to  or  less  than  %d,  the  diameter. 

The  volume  of  one  pressure  may  be  computed  from  Eq.  17, 


*  Senate  Ex.  Doc.  No.  1— 47th  Congress,  2d  Session. 


134 


STRENGTH   OF   BEAMS   AND    COLUMNS. 


page  16,  or  from  computing  and  tabulating  the  volumes  for  all 
required  values  of  dQ,  when  the  radius 
is  unity,  which  we  will  represent  by 
fC,  and  since  the  volumes  of  cylindrical 
wedges  are  as  the  squares  of  their  radii, 
we  have  for  the  required  load,  L, 


L  =  %r'fC. 


(100) 


Table  of  positions  of  the  neutral  line,  d^  and  the  computed 
values  of  f  for  the  various  required  values  of  q  =  C  -=-  T, 


?• 

C?e. 

/. 

Q. 

dc. 

/• 

8.0 

0.3984  d 

0.4877 

4.0 

0.4951  d 

0.6578 

7.875 

0.4004  d 

0.4911 

3.866 

0.5000  d 

0.6666 

7.75 

0.4022  d 

0.4942 

3.75 

0.5045  d 

0.6749 

7.625 

0.4041  d 

0.4975 

3.625 

0.5095  d 

0.6838 

7.5 

0.4062  d 

0.5011 

3.5 

0.5146  d 

0,6947 

7.375 

0.4087  d 

0.5054 

3.375 

0.5200  d 

0.7034 

7.25 

0.4112  d 

0.5097 

3.25 

0.5256  d 

0.7137 

7.125 

0.4137  d 

0.5139 

3.125 

0.5314  d 

0.7239 

7.0 

0.4161  d 

0.5180 

3.0 

0.5375  d 

0.7354 

6.875 

0.4186  d 

0.5221 

2,875 

0.5438  d 

0.7469 

6.75 

0.4209  d 

0.5265 

2.75 

0.5505  d 

0.7691 

6.625 

0.4236  d 

0.5311 

2.625 

0.5574  d 

0.7723 

6.5 

0.4262  d 

0.5355 

2.5 

0.5646  d 

0.7851 

6.375 

0.4289  d 

0.5403 

2.375 

0.5724  d 

0.7993 

6.25 

0.4316  d 

0.5446 

2.25 

0.5804  d 

0.8143 

6.125 

0.4344  d 

0.5496 

2  125 

0.5890  d 

0.8302 

6.0 

0.4373  d 

0.5549 

2.0 

0  5980  d 

0.8469 

5.875 

0.4402  d 

0.5599 

1.875 

0.6076  d 

0.8641 

5.75 

0.4432  d 

0.5652 

1.75 

0.6178  d 

0.8838 

5.625 

0.4463  d 

0.5606 

.625 

0.6277  d 

0.9042 

5.5 

0.4494  d 

0.5761 

.5 

0.6404  d 

0.9260 

5.375 

0.4526  d 

0.5819 

.375 

0.6525  d 

0.9496 

5.25 

0.4559  d 

0.5877 

.25 

0.6666  d 

0.9747 

5.125 

0.4594  d 

0.5937 

.125 

0.6816  d 

1.0039 

5.0 

0.4629  d 

0.6000 

.0 

0.6976  d 

1.0331 

4.875 

0.4665  d 

0.6066 

0.875 

0.7162  d 

1.0661 

4.75 

0.4702  d 

0.6132 

0.75 

0.7357  d 

1.1042 

4.625 

0.4740  d 

0.6199 

0.625 

0.7583  d 

1.1466 

4.5 

0.4780  d 

0.6270 

0.5 

0.7843  d 

1.1946 

4.375 

0.4821  d 

0.6343 

0.375 

0.8149  d 

1.2511 

4.25 

0.4863  rf 

0.6419 

0.25 

0     d 

4.125 

0.4906  d 

0.6496 

STRENGTH   OF   COLUMNS.  135 

EXAMPLE  48. — Required  the  breaking  load  of  a  Cylindrical 
Cast-Iron  Column,  when 

The  diameter  d  —     1".129,  C  =  96280  Ibs.  computed  in  Ex.  3, 
"    length.  ..!==   10".0,     T=  29400  "    by  test, 
"   deflection  3=     0".4,      ?=  3.271. 

The  value  of  f  from  the  Table  corresponding  to  this  value 
of  q  is  f  —  0. 715,  then  from  Eq.  100  we  have 

L  —  2  X  (0.5645)2  0.715  X  96280  =  43884  pounds. 

This  column  was  cast  from  the  iron  described  in  Example 
14,  page  82,  "  Cracks  developed  on  the  tension  side  when 
the  deflection  reached  0".4,  the  load  sustained  being  40000 
pounds."  Four  other  columns  of  the  same  iron  were  broken, 
with  deflections  varying  from  0".38  to  Ov.43,  but  the  loads 
sustained  were  not  given.  The  theoretical  determination  of 
the  position  of  the  neutral  line  and  the  practical  is  almost 
identical,  the  theoretical  being  dT  =  0.538,  and  the  practical 
dT  =  0".40. 

HOLLOW  CIRCULAR  PILLARS. 

For  this  Case  the  deflection,  #,  must  be  equal  to  or  greater 
than  6?T,  the  depth  of  the  tension  area  given  by  the  following 
Table  for  the  neutral  line,  when  q  = 
C  -f-  T  for  the  material,  and  equal  to  or 
less  than  \d,  the  diameter  of  the  outer 
surface  of  the  pillar. 

The  volume  of  one  pressure  wedge  may 
be  computed  from  Eq.  20,  page  20,  or 
from  computing  and  tabulating  the  vol- 
umes for  all  required  values  of  d^  when  the  radius  is  unity, 
which  we  will  represent  by  tf0O,  and  since  they  are  as  their 
radii  we  have  for  L  the  load,  t  being  the  thickness  of  the 
metal — 

L  =  2rtf0C.  (101) 


136 


STKENGTH    OF   BEAMS   AND    COLUMNS. 


Table  of  positions  of  the  neutral  line,  dc,  and  the  computed 
values  of  f0  for  the  various  required  values  of  q  =  C  -=-  T. 


Q- 

dc. 

/o- 

Q- 

dc. 

/o- 

8.3195 

0.5000  d 

2.0000 

4.5 

06520  d 

2.3377 

8.0 

0  5093  d 

2.0214 

4.375 

0.6590  d 

2.3528 

7.875 

0.5131  d 

2.0299 

4.25 

0  6662  d 

2.3684 

7.75 

0.5170  d 

2.0389 

4.125 

0.6735  d 

2.3845 

7.625 

0.5209  d 

2.0481 

4.0 

0.6811  d 

2.4010 

7.5 

0.5249  d 

2.0566 

3.875 

0.6889  d 

2.4180 

7.375 

0.5290  d 

2.0657 

3.75 

0.6970  d 

2.4355 

7.25 

0  5331  d 

2.0752 

3.625 

0.7052  d 

2.4533 

7.125 

0.5374  d 

2.0847 

3.5 

0.7137  d 

2.4717 

7.0 

0.5418  d 

2.0945 

3375 

0.7223  d 

2.4906 

6.875 

0.5462  d 

2.1044 

3.25 

0.7313  d 

2.5101 

6.75 

0.5507  d 

2.1144 

3125 

0.7405  d 

2.5301 

6.625 

0.5553  d 

2.1248 

3.0 

0.7500  d 

2.5509 

6.5 

0.5600  d 

2.1351 

2.875 

0.7597  d 

2.5722 

6.375 

0.5647  d 

2.1457 

2.75 

0.7697  d 

2.5941 

6.25 

0.5697  d 

2.1568 

2625 

0.7800  d 

2.6167 

6.125 

0.5748  d 

2.1680 

2.5 

0.7906  d 

2.6400 

6.0 

0.5800  d 

2.1790 

2.375 

0.8014  d 

2.6638 

5.875 

0.5852  d 

2.1910 

2.25 

0.8122  d 

2.6877 

5.75 

0.5907  d 

2.2035 

2.125 

0.8241  d 

2.7143 

5.625 

0.5962  d 

2.2151 

2.0 

0.8358  d 

2.7404 

5.5 

0.6018  d 

2.2277 

1.875 

0.8478  d 

2.7673 

5.375 

0.6076  d 

2.2104 

1,75 

0.8600  d 

2.7950 

5.25 

0.6134  d 

2.2530 

1.625 

0.8724  d 

2.8234 

5.125 

0.6193  d 

2.2663 

1.5 

0  8851  d 

2.8526 

5.0 

0.6255  d 

2.2798 

1.375 

0.8979  d 

2:8823 

4.875 

0.6314  d 

2.2933 

1.25 

0.9107  d 

2.9124 

4.75 

0.6384  d 

2.3079 

1.125 

0.9233  d 

2.9425 

4.625 

0.6451  d 

2.3226 

1.0 

0.9360  d 

2.9732 

EXAMPLE  49.— Required  the  Greatest  Strength  of  a  Phoenix 
Column,  when 

The  outer  diameter  ....d  =  8".00,       q  —  1, 
"    thickness  of  metal,  t  =  0".35,     /0  =  2.9732   for  q  =  1, 
"    deflection ..<?  =  0".535,     Z  =  7'.0. 

The  crushing  value  of   6"  =  60000  is  assumed,  then  from 
Eq.  101, 

L  =  2  X  4.0  X  0.35  X  2.9732  X  60000  =  499497  pounds. 


STRENGTH   OF   COLUMNS.  137 

By  the  United  States  Government  Watertown  Arsenal 
tests  the  greatest  strength  was  468000  pounds.* 

117.  Case  V.—  PILLARS    THAT   CROSS-BREAK    FROM   COM- 

PRESSION AND  CANTILEVERAGE. 

The  shortest  pillar  of  this  class,  for  a  given  section  and 
material,  is  that  length  that  just  deflects  sufficiently  to  allow  the 
resultant  of  the  applied  load  to  pass  to  the  right  of  the  fulcrum 
f  (Fig.  34).  This  for  a  given  material  varies  with  the  section, 
and  the  class  includes  all  pillars  of  longer  lengths.  The 
amount  of  direct  pressure  along  the  \mefn  is  equal  to  the  sum 
of  the  two  equal  pressure  wedges  described  in  Case  IY.  for 
the  given  cross-section,  and  is  the  same  for  all  deflections,  but 
to  obtain  the  breaking  load  of  the  pillar  this  sum  must  be  di- 
minished by  the  tensile  and  compressive  strain  that  arises  from 
the  load's  action  as  a  cantilever. 

C  '•  =  the  greatest  intensity  of  the  compressive  strain  arising 
from  bending  action  of  the  load. 

RECTANGULAR  PILLARS. 
d  —  \d  =  the  lever-arm  of  the  load,  L. 

Then  the  moment  of  the  applied  load  will  be,  from  Eq.  23, 
p.  38, 


r,  _  L  (W  -  d) 
~~MT 

Then  we  will  have  for  the  applied  load,  Z, 
L  =  MCC-MCC', 

*  Ex.  Doc.  No.  23,  House  of  Representatives,  46th  Congress,  3d  Session, 
p.  278. 


138  STRENGTH    OF    BEAMS   AND    COLUMNS. 

from  which,  by  substituting  the  above  value  of   C' ',  we  ob- 
tain 

•  r  _       M*C 

•L-    --- 


d 


When  3#  becomes  equal  to  or  less  than  ^,  the  depth,  the  for- 
mula becomes  that  for  rectangular  pillars 
of  Case  IV.,  the  cantilever  effect  of  the 
load  having  disappeared. 

As  before  stated,  the  formulas  deduced 
1       for  the  strength  of  columns  that  fail  by 
cross-breaking  with  cantileverage  are  iden- 
tical in  form   with  those  given  by  Mr. 
Lewis  Gordon,  but,  unlike  his,  they  give 

exact  values  and  show  when  the  formula  does  not  apply.  This 
uncertainty  as  to  the  length  of  pillars  to  which  Gordon's  for- 
mulas did  not  apply  has  been  the  chief  objection  to  their  use 
in  practice. 

Adapting  Gordon's  formula  for  the  strength  of  rectangu- 
lar pillars  to  our  notation,  we  have 

AC 


From  making  A  the  full  area  of  the  column,  the  factor  C 
could  never  be  the  crushing  strength  of  the  material,  but  is 
always  something  less  than  it  in  value.  The  second  member 
of  the  denominator  makes  the  formula  true  for  all  lengths  of 
pillars,  which  experiment  does  not  confirm,  and  is  of  the  same 
general  form  as  our  formula  for  the  deflection  given  by  Eq. 
93. 

~No  successful  effort  has  ever  been  made  to  so  modify  Gor- 
don's formula  that  the  computed  and  experimental  strength 


STRENGTH   OF   COLUMNS.  139 

of  wooden  columns  would  be  the  same  in  value  ;  and  for 
the  reason  that  in  all  published  results  of  experiments  on 
wooden  pillars,  except  for  a  few  on  yellow  pine  made  at 
the  Watertown  Arsenal,  they  all  failed  without  cantilev- 
erage  of  the  applied  load,  and  the  deflection  did  not  enter 
as  a  factor  to  decrease  the  strength  as  contemplated  by  Gor- 
don. 

In  the  tests  of  white  and  yellow  pine  columns  made  at  the 
Watertown  Arsenal,  and  described  in  Examples  46  and  47,  the 
strength  was  not  decreased  from  canti  leverage.  For  any 
given  rectangular  section  of  these  pillars,  the  strength  is  the 
same  for  all  lengths  from  twenty-seven  to  forty-five  times  the 
least  diameter  or  least  side  of  the  rectangle,  which  includes 
the  lengths  of  all  yellow  arid  w^hite  pine  pillars  that  are  used 
in  structures. 

EXAMPLE  50.  —  Required  the  breaking  load  of  a  Rectangu- 
lar Wrought-Iron  Column,  tested  with  two  V\  pin  ends, 
when 

The  diameter  .....  d  —  3".0,     C  =  50000  pounds  assumed. 

"     breadth  ......  I  —  3".0,     T  —  50000  pounds  mean  tests. 

"     deflection  ....£=  1"  .61,    q  =  I. 

The  position  of  the  neutral  line  from  the  Table,  page  92,  is 
dc  —  0.78^;  from  Eq.  102  we  have 


Z  =  8  X  ^  X  5QOO°  =  -  =  197000  pounds. 

3  X  1.61-3.0       1  -f  0.783 

2.34  " 


This  example  and  those  in  the  following  Table  are  from  the 
series  of  tests  described  in  Example  44,  page  128,  as  given  in 
"  Senate  Ex.  Doc.  No.  5—  48th  Congress,  1st  Session,"  pp.  68 
to  102. 


140  STRENGTH    OF   BEAMS   AND    COLUMNS. 

Dimensions  3"  X  3". 


DEFLECTION 

IN  INCHES. 

LOAD  IN 

POUNDS. 

Length  in 

/  -*•  d. 

Inches 

From. 

To. 

Computed. 

Experimental. 

28 

84 

0.5 

216000 

30 

90 

0.5 

218975 

32 

96 

0.5 

.61 

197000 

219000 

34 

102 

0.3 

61 

197000 

208500 

36 

108 

0.3 

*  '.80 

173760 

190500 

38 

114 

0.5 

.80 

173760 

181125 

40 

120 

0.6 

.9 

163000 

181750 

42 

126 

0.5 

.67 

188800 

169700 

44 

132 

0.4 

2.30 

131700 

178950 

46 

138 

0.4 

1.98 

155600 

157325 

48 

144 

0.45 

2.16 

141100 

155790 

50 

150 

0.45 

2.30 

131700 

153065 

52 

156 

0.35 

2.50 

120088 

155000 

54 

162 

0.30 

2.50 

120088 

147750 

56 

168 

0.30 

2.80 

106000 

149875 

58 

174 

0.35 

2.50 

120088 

128775 

60 

180 

0.30 

2.50 

120088 

126875 

When  under  the  greatest  load  sustained  by  the  above  col- 
umns, they  "  suddenly  sprung "  to  a  cross-'breaking  deflection 
with  cantileverage.  The  "  Deflection  from  "  in  the  Table  is 
that  at  which  the  "  sudden  spring "  began,  and  "Deflection 
to  "  is  that  to  which  it  sprung  /  the  deflection  in  the  Table 
is  the  mean  of  those  of  two  tests,  in  most  cases. 

The  "  Experimental "  load  is  the  mean  of  two  tests,  in  each 
case,  and  is  that  sustained  by  the  column  at  the  beginning  of 
the  "  sudden  spring."  From  the  "  Computed  "  loads  it  will 
be  seen  that  the  momentum  of  the  "  sudden  spring  "  caused 
the  observed  deflections  to  exceed  the  true  deflection  in  many 
examples  of  this  series  of  tests. 

CIRCULAR  PILLARS. 

d  —  f d  =  the  lever-arm  of  the  load,  Z, 
f  =  the  factor  in  Table,  page  134, 
/c  =  the  «  «  «  pages  80,  100  and  112. 


STRENGTH    OF   COLUMNS. 


141 


From  Eq.  76,  page  57,  the  moment  of  the  applied  load,  Z, 
becomes 

L  (8  -  \d)  =  r>feC", 


__  L  (S3  —  3d) 


and  for  the  load  from  the  pressure  wedges  given  by  Eq.  100 
we  have 

L  =  WfC—  2/y<7'. 


Substituting  for  C'  its  value  given  above,  we  have 


(103) 


When  8#  becomes  equal  to  or  less  than 
3d  the  load  must  be  computed  from  the 
formula  for  Circular  Pillars,  Case  IY., 
as  in  this  position  of  the  pillar  there  will 
be  no  cantilever  effect  of  the  applied 
load  to  be  deducted  from  the  sum  of  the 
two  pressure  wedges. 


HOLLOW  CIKCULAK  PILLARS. 

d  =  the  outer  diameter  in  inches, 

r  =  the     "      radius       "       " 

=  the  lever-arm  of  the  load,  Z, 
fo  =  the  factor  given  in  Table,  page  136, 
/c  =  the     "  "  "       pages  85  and  104. 

t  =  the  thickness  of  the  metal  ring  in  inches. 


From  Eq.  91,  page  62,  the  moment  of  the  applied  load  be- 


comes 


142 


STRENGTH    OF   BEAMS   AND   COLUMNS. 


,- 

"s^r 

For  the  load,  Z,  we  have 

L  =  2rtf0C  -  2rtfQCf 
Substituting  the  value  of  C',  we  obtain 

T 
~ 


1    , 


(104) 


2/tf 


When 


is  equal  to  or  less  than  the  outer  diameter,  d,  the 
strength  of  the  column  must  be  com- 
puted from  Eq.  101,  Case  IV.,  as  there 
will  then  be  no  cantilever  effect  to  de- 
duct. 


EXAMPLE  51.  —  Required  the  Breaking 
Strength   of  a   Phoenix   Column  tested 
with  flat  ends  ;    C  =  T  =  60000  is  the 
assumed  strength  of  the  iron. 

Outer  diameter  ____  d  —  8".0,  q   =  C  -4-  T  —  1, 

Thickness  of  metal,  t  =  0".35,  /o  =  2.9732  from  Table,  page  136,  for  q  =  l, 

Deflection  .........  d  =  2".47,/c  =  2.6806     "         "  "     104,    "    q  =  l. 

From  Eq.  104  we  have 

2  X  4  X  0.35  X  2.9732  X  60000 


T  _ 


499497 


,   2.9732  (4  X  2.47 
2  X  4  X  2.6806 


-  8) 


1+0.26 


This  example  is  taken  from  the  series  of  experiments*  de- 
scribed in  Example  49,  page  136  ;  the  tested  breaking  load  was 
416000  pounds  ;  the  length  was  28  feet. 

*  Report  of  the  United  States  Board  appointed  to  test  iron  and  steel,  Ex. 
Doc.  No.  23,  House  of  Representatives,  46th  Congress,  2d  Session,  page 
270. 


STRENGTH   OF   COLUMNS.  143 

ANGLE-IRON,  Box,  CHANNEL,  EYE-BEAM  AND  TEE  PILLARS. 

In  pillars  of  the  above  sections,  the  distance  of  the  neutral 
line  from  that  side  of  the  pillar  that  will  most  likely  be  its 
concave  side,  when  broken,  must  be  computed  by  the  rules 
heretofore  given  for  it  in  beams  of  the  section  of  the  pillar. 
The  sum  of  the  two  pressure  wedges  will  then  be  the  product 
of  the  compressed  area,  A^  by  the  crushing  strength  of  the 
material. 

The  distance,  g,  of  the  centre  of  gravity  of  the  applied 
load  wedge  from  the  point  k  (Fig.  34),  must  be  determined, 
which  is  obtained  by  computing  RF,  the  moment  of  the  load 
wedge  with  respect  to  the  axis,  &,  in  which  F  is  the  greatest 
intensity  of  the  load  at  &,  it  being  zero  at  w,  and  It  a  factor 
that  depends  for  its  value  upon  the  section  ;  for  any  given 
dimensions  it  reduces  to  a  numerical  quantity  without  assign- 
ing any  value  to  F\  and  dividing  this  moment,  RF\  by  the 
volume  of  the  load  wedge,  AF  -r-  2,  in  which  A  is  the  area 
of  the  section,  we  have 


and 


Then  we  have 

T  -       AH          AT' 

JLy     -J-l-c^ J-J-c^     • 

Substituting  the  above  value  of  (7',  we  deduce 

ACO 
~-  1  +  A(*  --<?).  (105) 

7?c  is  a  factor  of  the  moment  of  resistance  that  the  section 
of  the  pillar  offers  to  the  cantilever  bending  of  the  load  ;  for 
a  given  section  it  becomes  a  numerical  quantity  without  assign- 


144  STRENGTH   OF  BEAMS    AND    COLUMNS. 

ing  any  value  to  C'  when  computed  by  the  rules  given  for  the 
moment  of  resistance  of  the  section  when  strained  in  a  beam. 
When  d  becomes  equal  to  or  less  than  g  in  value  the  canti- 
lever strain  ceases  to  exist,  and  the  pillar  belongs  to  Case  IV., 

.-.  L  =  A6C.  (106) 

When  6  becomes  less  than  dT,  the  depth   of  the  extended 
area  required  for  rupture,  the  pillar  belongs  to  Case  III. 


in  which  dT  gives  the  position  of  the  neutral  line  of  rupture, 
and  <i  the  ratio  of  the  compressive  strain  that  will  be  required 
to  hold  in  equilibrium  the  tensile  strain  developed  by  the 
bending  of  the  pillar  as  a  cantilever. 

Equation  105  is  the  general  formula  for  the  strength  of 
pillars  of  all  lengths,  sections  and  material  of  which  the 
formulas  heretofore  deduced  in  this  chapter  are  only  the  forms 
it  will  assume  for  special  cases.  The  denominator  of  the 
second  member  of  the  formula  must  never  be  less  than  unity. 


CHAPTEE   VIII. 

COMBINED  BEAMS  AND   COLUMNS. 

118-  General  Statement.  In  roof -trusses,  cranes, 
derricks,  platforms  supported  bj  cantilevers,  trussed  beams  and 
other  structures,  there  is  used  a  class  of  pieces  of  material  that, 
from  the  manner  in  which  they  are  loaded,  do  not  belong  ex- 
clusively to  either  horizontal  beams  or  columns,  but  partake  of 
the  nature  of  both,  in  the  manner  in  which  they  support  the 
load  to  which  they  are  subjected. 

The  theory  of  the  transverse  strength  of  these  Combined 
Beams  and  Columns  gives  the  solution  of  the  general  problem 
of  the  transverse  strength  of  all  beams,  without  regard  to  the 
special  angle  that  the  axis  of  the  beam  makes  with  the  line  of 
direction  of  the  loading  and  supporting  forces.  Horizontal 
beams  acting  under  vertical  loads  and  columns  are  only  special 
cases  of  the  general  problem,  in  which  certain  factors,  that 
cause  the  strength  of  the  same  piece  of  material  to  vary  with 
its  angle  of  inclination  to  the  horizon,  disappear,  from  the 
general  rule  for  these  cases,  by  becoming  zero  in  value. 
But  on  account  of  their  great  importance,  we  have,  in  the 
preceding  chapters,  deduced  separately  the  principles  and  rules 
from  which  the  strength  of  these  special  cases  may  be  com- 
puted. 

.  Should  one  end  of  a  horizontal  beam,  such  as  be  (Fig.  44), 
be  fixed  in  a  vertical  wall,  and  a  load  be  attached  to  its  free 
end,  not  suspended,  as  in  the  figure,  none  of  this  load  will  di- 
rectly compress  or  rest  upon  the  cross-section  of  the  beam,  ~be. 
ISTow  let  the  wall  be  revolved  around  the  point  F,  to  a  horizon- 
tal position,  thus  bringing  the  beam,  ~be,  to  the  vertical,  then  the 


146  STRENGTH    OF   BEAMS   AND   COLUMNS. 

entire  load  will  rest  upon  or  directly  compress  the  cross-section 
of  the  beam,  now  converted  into  a  pillar ;  the  load  has  thus  been 
gradually  converted  from  a  non-compressing  to  a  compressing 
load  with  its  full  weight.  The  bending  moment  of  the  attached 
load  is  greatest  when  the  beam  is  in  the  horizontal  position,  and 
it  gradually  diminishes  as  its  lever-arm,  ,<?,  becomes  less  in  value 
with  the  revolution  of  the  wall,  and  becomes  zero  in  value 
when  the  beam  occupies  the  vertical  position.  On  the  other 
hand,  should  the  wall  be  revolved  to  the  horizontal  position 
around  the  point  P,  the  bending  moment  will  gradually  de- 
crease and  become  zero,  while  the  tensile  strain  will  increase 
from  zero  to  that  of  the  full  weight  of  the  load,  when  the 
beam  becomes  vertical. 

From  the  above  illustration  the  origin  of  the  special  cases 
of  horizontal  beams  and  columns  is  apparent,  and  the  reason 
for  the  special  rules  for  their  strength.  A  similar  illustration 
could  be  deduced  from  a  beam  supported  at  both  ends,  by  con- 
ceiving it  to  occupy  all  positions  from  the  horizontal  to  the 
vertical. 

In  order  to  deduce  the  relation  that  exists  between  the 
applied  load  and  the  resistance  of  the  combined  beam  and 
column  at  the  instant  of  rupture,  it  is  assiimed,  in  the  analy- 
sis, that  they  only  deflect  enough  to  admit  of  their  cross-break- 
ing as  a  column,  without  cantileverage  of  the  applied  load,  as 
in  Class  IV.,  Chapter  VII.  In  very  long  beams,  however,  the 
compression  resulting  from  pressure  applied  to  its  ends  will 
act  with  the  leverage  explained  in  Case  V.,  page  137,  and  the 
strength  must  be  computed  from  the  formulas  there  given,  but 
the  compression  resulting  from  the  transverse  bending  of  the 
load  will  be  the  same  in  each  case,  and  will  compress  the  beam 
without  cantileverage. 

119.  Notation.  In  addition  to  the  notation  heretofore 
used  and  defined  in  Art.  35,  page  37,  the  following  will  be 
used  in  this  Chapter  : 


COMBINED    BEAMS    AND    COLUMNS.  147 

C  =  6"+  C"  —  the  greatest  intensity  of  the  compressive  strain 

in  pounds  per  square  inch, 

C'  —  the  greatest  intensity  of  the  compressive  strain  in  pounds 
per  square  inch,  arising  from  the  bending  component 
of  the  load, 

<7"  —  the  greatest  intensity  of  the  compressive  strain  in  pounds 
per  square  inch,  arising  from  the  compressing  compo- 
nent of  the  applied  load, 

ZB  —  the  bending  component  of  the  applied  load  in  pounds, 
Zc  ==  the  compressing  component  of  the  applied  load  in  pounds, 
I  =  the  unsupported  length  of  the  beam  in  inches, 
8  =  the  span,  the  horizontal  distance  between  the  supports 

in  inches, 
h  ~  the  difference  between  the  heights  of  the   ends  of  the 

beam  in  inches, 
a  =  the  angle  that  the  beam  makes  with  the  horizon. 

In  many  of  the  different  methods  of  loading  and  supports 
ing  beams  given  in  this  Chapter,  the  greatest  resulting  com- 
pressive strain  at  any  section  is  the  sum  of  two  or  more  dis- 
tinct pressures,  which  will  be  represented  by  the  letter  C,  with 
corresponding  accents,  such  as  C',  C",  C'",  C"" ,  etc. 

INCLINED  BEAMS. 

12O.  General  Conditions.  The  effect  produced  by 
a  load  when  applied  to  this  class  of  beams  will  manifest  itself 
in  two  distinctly  different  ways  ;  each  separate  effect  must  be 
computed,  and  their  sum  will  be  the  total  effect  produced  -by 
the  load  upon  the  beam.  This  is  accomplished  by  decompos- 
ing the  applied  load  into  two  components,  by  the  well-known 
theorem  of  the  parallelogram  of  forces.  One  component,  ZB, 
must  be  at  right  angles  to  the  axis  of  the  beam,  and  the  other, 
Zc,  parallel  with  it ;  the  first  component  will  bend  the  piece  as 
a  learn,  while  the  second  will  compress  it  as  a  column. 

The  parallelogram  of  forces  for  inclined  beams  is  a  rect- 


148 


STRENGTH   OF   BEAMS   AND   COLUMNS. 


angle,  and  the  relation  between  the  components  and  the  load 
is  obtained  from  that  of  the  three  sides  in  a  right-angle  tri- 
angle, in  which 


cos.  a~  — 

V 


sin.  a  =-  — 


tan.  a  =  — . 


121.  Inclined  Beam  Fixed  and  Supported  at 
One  End.  This  method  of  "  fixing  "  and  loading  beams  is 
illustrated  by  two  different  positions  of  the  beam,  &<?,  in  Fig. 
44.  Two  Cases  will  be  considered. 

Case  I.—  When  the  load  is  applied  at    the  free  end  of 

the  beam. 

Let  the  right-angle  triangle,  eot  (Fig.  44),  represent  the 
parallelogram  of  forces,  •  in  which  et  represents  the  applied 


load,  drawn  to  any  given  scale,  then  ot  will  be  the  lending  and 
oe  the  compressing  component  of  the  applied  load,  Z,  from 
which 

ot  —  et  cos.  a  and  oe  —  et  sin.  a, 

T     _Ls 

'  '  ^  ~  T' 


(108) 


COMBINED   BEAMS   AND   COLUMNS.  149 

and 

L,  =  ~-  (109) 

The  bending  moment  of  the  component,  ZB,  its  lever-arm 
being  £,  will  be 

/«? 

Bending  Moment,  ZB  ==  -,-X  I  —  Ls.  (110) 

i 

Having  decomposed  the  applied  load,  Z,  into  two  compo- 
nents, one  perpendicular  and  the  other  parallel  to  the  inclined 
beam,  we  can  now  ascertain  the  effect  that  will  be  produced  by 
the  original  load  upon  it,  by  a  combination  of  the  methods 
used  to  compute  the  effect  produced  upon  horizontal  beams 
and  columns  by  vertical  applied  loads. 

RECTANGULAR  BEAMS. 

The  transverse  strength  of  a  rectangular  beam  loaded  and 
fixed  as  in  this  Case  will  now  be  deduced  from  the  foregoing 
formulas. 

The  bending  moment  from  Eq.  110  must  be  made  equal  to 
the  moment  of  resistance  from  Eq.  23,  page  38,  in  which 
C=  C'. 

...Z*=^',  (111) 

3  Te 

.>.'•  f±?=±  U,C'.  (112) 

a 


c 


The  compressing  component,  Zc,  from  Eq.  109,  must  be  re- 
sisted by  an  equal  compression  produced  in  the  section  at  the 
face  of  the  wall,  as  given  by  Eq.  99,  page  131,  in  which 

C  =  C"  and  L  =  ^, 
l> 

*=U,C".  (113) 


150  STRENGTH    OF   BEAMS   AND    COLUMNS. 

Adding  Eqs.  112  and  113  we  have,  by  making  C  '  +  C"  — 
Cj  the  crushing  strength  of  the  material, 


from  which 

r  _ 

- 


When  the  beam  is  horizontal,  h  —  0  and  I  =  s,  the  formula 
reduces  under  this  hypothesis  to  that  given  in  Eq.  27,  page  38, 
in  which  m  —  1.  In  a  vertical  beam,  s  =  0  and  I  —  A,  the 
formula  reduces  to  that  given  for  the  strength  of  rectangular 
columns,  Eq.  99. 

To  Design  a  Rectangular  Beam. 

The  length  I  and  height  h  will  be  controlled  by  the  position 
in  which  the  beam  is  to  be  used.  A  convenient  depth,  d, 
must  then  be  assumed,  and  the  value  of  dc  computed  from 
Eq.  26,  page  38  ;  then  from  Eq.  114  we  obtain 


a     L 

~id?e-  > 

from  which  the  required  breadth,  J,  of  a  beam  that  will 
break  with  a  given  load,  Z,  will  be  obtained  by  giving  to  C 
the  value  of  the  crushing  strength  of  the  material  of  which 
the  beam  is  to  be  constructed. 

Case   II.  —  WHEN    THE   LOAD    is    UNIFORMLY  DISTRIBUTED 

OVER  THE  UNSUPPORTED  LENGTH  OF  THE  BEAM. 

The  resultant  of  the  applied  load  will  pass  through  the 
middle  of  the  length  of  the  beam  and  the  triangle,  eot  (Fig. 
44),  will,  as  in  Case  I.,  give  the  relation  between  the  load  and 
its  two  components  ;  hence 


(118) 


COMBINED   BEAMS    AND    COLUMNS.  151 

The  bending  moment  of  the  component,  ZB,  will  be, 

Bending  Moment  :=—-><  —  —   — -,  (H^) 

£         2          2 

its  lever-arm  being  I  -=-  2. 

RECTANGULAR  BEAMS. 

The  transverse  strength  of  a  rectangular  beam  loaded  and 
fixed  as  in  this  Case  may  be  obtained  by  the  same  process 
heretofore  used  in  Case  L, 

(120) 


To  Design  a  Rectangular  Beam. 

Assume  a  depth,  d,  and  compute  dc  from  Eq.  26,  page  38, 
then  from  Eq.  120  we  have 


±22.  Inclined  Beam,  supported  at  one  end 
and  STAYED  or  held  in  position  at  the  other  with- 
out vertical  support.  This  class  of  Inclined  Beams  is 
illustrated  in  Figs.  45,  46  and  47.  Five  different  cases  of 
loading  will  be  considered. 

Case  I.  —  WHEN   THE  INCLINED  BEAM   is    LOADED  AT    ITS 

STAYED    END. 

The  effect  produced  by  the  load,  L,  upon  the  beam,  be  (Fig. 
45),  is  simply  to  compress  it  as  a  column,  the  load  being  held 
in  equilibrium  by  a  pull  along  the  tie  ec,  and  a  thrust  along 
the  inclined  beam. 

In  the  right-angle    triangle,  toe,  let  et  represent  the  load 


152 


STRENGTH  OF  BEAMS  AND  COLUMNS. 


drawn  to  any  given  scale,  then  to  gives  the  pull  and  oe  the 
thrust. 

et  te 


ot  = 


tan.  a 


,    oe  =  — 


s^n.  a 


and 


Ls 

The  pull  =  -7-5 

The  thrust  Lc  =  -v-? 


(122) 


(123) 


from   which  the  compressing  effect  of  the  applied  load,  Z, 
can  be  ascertained. 

RECTANGULAR  BEAMS. 

The  direct  compression  given  by  Eq.  123  must  be  made 
equal  to  the  resistance  offered  by  the  section  of  the  beam  as  a 
column,  then  from  Eq.  99,  page  131, 

LI 

T  ~~  UA 

•.Z  =  ***£  (124) 


COMBINED    BEAMS    AND    COLUMNS. 


Case  II, — WHEN  THE  INCLINED  BEAM    is  LOADED  AT  ITS 
MIDDLE. 

Let  be  (Fig.  46)  represent  the  beam  loaded  at  its  middle,  t, 
with  the  load,  Z,  then  in  the  right-angle  triangle,  toe,  et  rep- 


resents the  load  drawn  to  any  scale,  ot,  perpendicular  to  the 
beam,  be,  the  lending  component,  ZB,  and  ot  parallel  to  be,  the 
directly  compressing  component,  Zc. 


Lh 
I  ' 


(125) 


(126) 


The  bending  moment  produced  by  the  component,  ZB,  is 
identical  with  that  produced  by  conceiving  the  beam  to  be 
"  fixed "  at  its  middle,  t,  and  loaded  at  the  free  end  with 
ZB  -f-  2, 

.  • .  Bending  Moment,  ZB  =  — 'T  X  ^  =  -7--         (127) 

2ii  2i  4: 

In  order  that  the  loading  and  supporting  forces  shall  be  in 
equilibrium,  one  half  of  the  bending  component,  ZB,  must  be 
supported  at  each  end  of  the  beam  b  and  c.  The  half  of  ZB 
at  c  cannot  be  directly  supported  at  that  point,  but  must  be 


154  STRENGTH   OF   BEAMS    AND    COLUMNS. 

carried  to  the  ground  by  some  means.  The  method  of  stay- 
ing the  end  of  the  beam,  be,  represented  in  the  figure  is  that 
used  in  roof -trusses ;  hence  the  load,  ZB  -j-  2  at  c,  must  be  de- 
composed into  two  components,  one  in  the  direction  of  each 
beam  or  rafter ;  if  the  angle,  bek,  is  a  right  angle  the  entire 
load,  ZB  -f-  2,  compresses  the  rafter,  ck,  as  a  pillar,  thus  reach- 
ing the  support,  k. 

In  roof -trusses  the  inclined  beams,  he  and  ck>,  are  usually 
loaded  in  the  same  manner,  and  each  beam  will,  therefore, 
carry  to  c  a  component,  ZB  -7-  2,  of  its  load  for  support,  which 
will  be  equivalent  to  the  component,  ZB  -4-  2,  compressing 
each  rafter  to  which  it  is  applied,  but  much  increased  in 
amount  from  the  manner  in  which  it  is  converted  from  a  load 
that  is  perpendicular  to  the  beam  into  a  compressing  load 
that  is  parallel  to  its  axis. 

In  the  two  equal  triangles,  cmn,  let  cm  in  each  represent  the 
load,  LB  -r-  2,  en.  the  equal  components  that  the  rafters  exchange 
with  each  other,  and  mn  the  equal  components  that  the 
loads,  ZB  -f-  2,  produce  that  strain  the  rafters  to  which  they 
are  applied.  When  the  angle,  tick,  is  greater  than  90°  the 
sum  of  the  components  (mn  -f-  en)  will  be  greater  than  ZB  -f-  2, 
when  ~bek  —  90°,  (mn  -j-  en)  =  ZB  -r-  2,  mn  becoming  zero. 
When  hcJc  is  less  than  90°  the  component,  mn,  is  a  tensile 
strain  ;  the  total  compression  (cm  —  mn)  will  then  be  less  than 

z.-a. 

In  roof -trusses  the  angle,  ~bck,  is  generally  greater  than  90°, 
and  the  total  compression  upon  each  rafter  resulting  from  the 
component  of  the  load  that  it  carries  to  c  is  (mn  -[-  en). 

mn  =  me  .  tan.  (90°  —  2«), 
me 


en,  = 


cos.  (90°  — 
21 


Ls 

Adding  and  substituting  me  =  —r,  we  have 


COMBINED    BEAMS   AND    COLUMNS.  155 

.  (128) 


The  factor,  c,  in  this  equation,  for  convenience  is 

c  =  tan.  (90°—  20)  -| — 5 .  (129) 

When  the  angle  a  =  45°,  tan.  (90°—  2a)  —  0, — 4 =  *> 

and  the  amount  of  compression  will  become  —  =  _      the 

2          21 
angle,  bck,  being  a  right-angle. 

The  total  compression  produced  by  the  single  load,  Z,  at  the 
middle  of  each  rafter  will  be  the  sum  of  the  three  distinct  parts, 
represented  by  Eqs.  126,  127  and  128. 

RECTANGULAR  BEAMS. 

The  amount  of  direct  compression  from  Eqs.  126  and  128 
must  be  equal  to  the  resistance  of  the  beam  as  a  column,  from 
Eq.  99,  page  131,  in  which  C  —  C'  and  C—  C'"  respectively. 

.-.    ^  =  bdcC",  (130) 

and 

^  =  bdeC'".  (131) 

The  bending  moment  from  Eq.  127  must  be  equal  to  the 
moment  of  resistance  of  the  section  as  a  beam  from  Eq.  23, 
page  38,  in  which  C  —  C". 

.'.     2L=M&,  (132) 

=  bdcC".  (133) 

Adding  these  separate  components,  as  given  by  Eqs.  130, 


156  STRENGTH    OF   BEAMS   AND    COLUMNS. 

131  and  133,  and  making  C'+  C"  +  C'"  =  C,  the  greatest 
compressive  strength  of  the  material,  we  have 


.     L_ 


from  which  the  required  breaking  load  may  be  computed. 
Case  III,  —  WHEN  THE    LOAD  is  UNIFORMLY   DISTRIBUTED 

OVER   THE    LENGTH  OF    THE    INCLINED    BEAM. 

The  bending  moment  and  the  component  Zc  will  be  identi- 
cal with  those  produced  by  conceiving  one  half  of  the  total 
load,  Z,  to  be  concentrated  at  its  middle  section,  t  (Fig.  46)  ; 
then,  from  Eqs.  125,  126  and  127,  Case  II.,  by  making 
L  =  L  -r-  2,  we  have 


^  =  ~,  (137) 

av 

Bending  Moment,  LK—  —£-•  (13  8) 

The  bending  moment  is  produced  by  a  component  of  the 
load,  Z,  that  is  perpendicular  to  the  length  of  the  beam,  and 
uniformly  distributed  over  its  length,  bo.  In  order  that  equili- 
brium shall  exist,  one  half  of  this  uniformly  distributed  com- 

Ls 

ponent,  __  must  be  supported  at  each  end  of  the  beam,  and  as 
L 

it  cannot  be  directly  supported  at  the  stayed  end,  c9,  it  must  be 
carried  to  the  supports  b  and  &,  as  described  in  Case  II.,  Eq. 
128. 

The  method  of  analysis  of  the  strains  in  a  simple  roof-truss, 
given  in  this  and  the  preceding  Case,  differs  from  that  usually 


COMBINED   BEAMS   AND   COLUMNS.  157 

pursued  by  writers.  One  half  of  the  load  on  each  rafter  is 
usually  considered  to  be  supported  by  b  and  k,  and  the  other 
half  of  each  rafter  load  to  be  supported  at  c9  by  reacting  against 
each  other.  The  transverse  strength  is  then  computed  as  if  the 
rafter  were  a  horizontal  beam,  which  is  equivalent  to  saying, 
that  for  a  given  material,  section  and  span  of  roof,  as  in  Fig. 
46,  the  transverse  strength  would  be  the  same  for  the  infinite 
number  of  roof  -trusses  that  could  be  constructed  between  these 
points  of  support  by  varying  the  pitch  or  angle  that  the  rafter 
makes  with  the  horizon,  which,  of  course,  is  not  the  case. 

RECTANGULAR  BEAMS. 


From  Eqs.  J28,  137  and  138,  we  obtain,  by  a  pi 
to  that  used  in  Case  III., 


(139) 


EXAMPLE  52.  —  Required  the  uniformly  distributed  breaking 
load  of  a  White  Pine  Rafter,  as  in  Fig.  46,  when 

The  span  ......  s  =  15'.  0,  C  =    5000  pounds  mean  of   tests, 

"    length  .....  I—  16'.  8,  T  =  10000       "  "       "       " 

"    rise  .......  h=    T.  5,   g=  0.5, 

"    depth  .....  d=    9".0,  de  =  OMSd,  from  Table,  page  108, 

"    breadth.  .  .b  =    5".0,  a  =-  26°.34'. 

From  Eq.  129,  c  =  2,  then,  from  Eq.  140, 


L  = 


X  201.6  X  (7.81  2)a  X  5000 


3X180X201.6+(4X90+2X2X38).S6SX9 


This  Example  is  taken  from  Trautwine's  "Engineer's 
Pocket-Book,"  and  was  designed  to  sustain  an  uniformly  dis- 
tributed load  of  8000  pounds,  with  a  factor  of  safety  of  three. 


158 


STRENGTH  OF  BEAMS  AND  COLUMNS. 


Case  IV. — WHEN  THE  INCLINED  BEAM  is  LOADED  AT  ITS 

MIDDLE,  AND  WITH  AN  ADDITIONAL  LoAD  AT  ITS  STAYED  END. 

The  bending  moment  and  the  compression  produced  by  the 
load,  Z,  applied  at  t,  the  middle  of  the  beam,  be  (Fig.  47),  will 


be  obtained  from  Eqs.  126,  127,  and  128  of  Case  II.,  the  two 
conditions  of  loading  being  the  same. 

One  half  of  the  additional  load,  ZA,  suspended  from  the 
stayed  end,  c,  is  supported  by  each  support,  b  arid  k ;  the  com- 
pression produced  upon  the  inclined  beam  as  a  column  has 
been  deduced  in  Case  I.,  Eq.  123;  the  horizontal  components, 
ZTand  F,  acting  in  opposite  directions,  neutralize  each  other. 

From    Eqs.  122  and  123,   page  152,  we  have,  by  making 


Horizontal  component,  II  =  -—-, 


The  Thrust,  Zc  =       -. 
2A 


RECTANGULAR  BEAMS. 


(141) 


(142) 


The  amount  of  direct  compression  from  Eqs.  126,  128  and 
142  must  be  made  equal  to  the  resistance  of  the  beam  as  a  col- 
umn, as  given  in  Eq.  99, 


COMBINED   BEAMS    AND   COLUMNS.  159 

**  =  M0C'9  (143) 

*™  =  1>dGC"'9  (144) 

2t 

LJ  =  ldcC"".  (145) 

Ail/ 

The  lending  moment  from  Eq.  127  must  be  equal  to  the 
moment  of  resistance  of  the  section  of  a  beam  from  Eq.  23, 
page  38,  in  which  C=  C", 


(147) 


Adding  Eqs.  143,  144,  145  and  147,  and  making 
C'  +  6r//  +  C'"+  C""  =  C,  the  crushing  strength  of  the  ma- 
terial, we  have 


r__ 


from  which  the  centre  breaking  load  of  the  beam  may  be  com- 
puted, in  addition  to  that  of  ZA  suspended  directly  from  its 
stayed  end. 

Case   V.  —  WHEN    THE  LOAD    is    UNIFORMLY    DISTRIBUTED 

OVER  THE  LENGTH  OF  THE  INCLINED  BEAM,  AND  AN  ADDITIONAL 
LOAD,  _Z/A,  APPLIED  TO  ITS  STAYED  END. 

This  Case  is  Cases  III.  and  IT.  combined,  therefore  Eqs. 
128,  137,  138  and  142  will  be  the  formulas  required  for 
computing  the  transverse  strength  of  the  beam. 


160  STEENGTH    OF   BEAMS    AND    COLUMNS. 

RECTANGULAR  BEAMS. 

The  direct  compression  from  Eqs.  128,  137  and  142  must 
be  equal  to  the  resistance  of  the  beam,  as  a  column,  from  Eq. 
99,  page  131. 

==  M0C',  (150) 


(151) 

£JIC 

Lcs 

±2  =  UJ3"".  (152) 

The  bending  moment  from  Eq.  138  must  be  equal  to  the 
moment  of  resistance  of  the  section  as  a  beam  from  Eq.  23, 
page  38,  in  which  C  —  C", 

£  =  ^',      '  (^3) 

~  ==  UJO".  (154) 

Adding   Eqs.    150,  151   and    152,    we   obtain   by   making 

C'  +  C"  +  C'"  +  C""  —  C. 


Inclined  Beam   Supported  vertically  at 
both  ends  and  loaded  at  its  middle. 

Let  Fig.  49  represent  the  beam,  supports  and  the  load,  Z. 
In  the  triangle,  toe,  let  eo,  to  any  scale  represent  the  load,  Z, 
then  the  component,  ot,  at  right  angles  to  the  beam,  will  bend 
it  and  et  parallel  to  the  beam  will  compress  it  as  a  column. 


COMBINED    BEAMS    AND    COLUMNS. 


161 


and 


Fig.  49. 

ot  =  eo  cos.  a  and  ot  —  eo  sin.  a, 
Ls 

^B   -=-j    , 

T        Lh 
A=   T- 


(157) 


(158) 


One  half  of  the  bending  component,  ZB,  will  be  supported  by 
the  walls  under  each  end  of  the  beam ;  the  bending  moment 
will  therefore  be 

Ls       I        Ls 
Sending  Moment,  ZB  =  — y  X  3-  =  (1^9) 

Li  2i  4: 

Having  decomposed  the  load  into  two  components,  one 
perpendicular  and  the  other  parallel  to  the  axis  of  the  inclined 
beam,  we  can  compute  its  effect  upon  a  beam  of  any  section 
and  material. 

LOAD  SUSTAINED  BY  THE  SUPPORTS. 

Each  of  the  walls,  supporting  the  ends  of  the  inclined  beam, 
must  resist  a  thrust  of  one  half  of  the  bending  component,  Lw 
which  tends  to  overturn  them.  Decomposing  this  thrust  at 
each  support  into  a  vertical  and  horizontal  component,  the 
first  will  be  the  proportion  of  the  centre  load,  Z,  that  is  sus- 
tained by  the  higher  support.  The  lower  support  also  sus- 


162  STRENGTH   OF   BEAMS   AND    COLUMNS. 

tains  an  equal  vertical  component,  and  in  addition  a  vertical 
component  of  the  compressing  component  of  the  applied 
load,  Z. 

The  horizontal  thrust  against  the  higher  support  will  be  the 
horizontal  component  of  one  half  of  ZB ;  that  at  the  lower 
support  will  be  the  sum  of  the  two  horizontal  components 
arising  from  decomposing  Zc  and  ZB  -7-  2  at  the  lower  support. 

From  the  above  we  obtain, 

Ls 

Vertical  load  on  higher  support  =  =  —j  •  cos.  a, 

L  f  s  \ 

Vertical  load  on  lower  support  =  -j  (  —  cos.  a  -{-  h  sin.  a\. 

When  the  beam  becomes  horizontal  the  angle  a  =  o,  s  =  I, 
cos.  a  =  1  and  the  sin.  a  =  o  ;  substituting  the  values  in 
the  above  formulas,  we  find  that  each  wall  supports  one  half 
of  the  load,  Z,  as  in  Case  III.,  page  4. 

RECTANGULAR  BEAMS. 

The  direct  compression  from  Eq.  158  must  be  equal  to  the 
resistance  of  the  beam  as  a  column  from  Eq.  99,  page  131. 

^  =  MCC'.  (160) 

L 

The  lending  moment,  from  Eq.  159,  must  be  equal  to  the 
moment  of  resistance  of  the  beam,  from  Eq.  23,  page  38. 

Ls  =  M^ 

4:  O 

.  - .  g*  -  ld.C*.  (162) 

Adding  Eqs.  160  and  161  we  have,  by  making  C'  +  C"  —  C, 
the  compressive  strength  of  the  material. 


COMBINED    BEAMS    AND    COLUMNS.  163 

(163) 
(164) 


from    which   the   centre  breaking   load    of    any    rectangular 
inclined  beam  supported  at  both  ends  may  be  computed. 


.  Inclined  Beams  supported  at  both  ends 
and  the  load  uniformly  distributed  over  its 
length. 

The  bending  moment  and  compression  are  identical  with  that 
produced  by  conceiving  the  beam  to  be  "  fixed"  at  its  middle 
and  loaded  on  its  free  ends  with  an  uniformly  distributed  load, 
L  -i-  2,  as  in  Case  II.,  page  150,  L  being  equal  to  Z  -=-  2 

Ls      .  Lh 

- 


The  bending  moment  will  be, 

Sending  Moment,  ZB  =  —  j  X  j-  =  -Q-  • 

21        4:         8 

From  these  components  the  effect  of  the  uniformly  distrib- 
uted load  on  an  inclined  beam  of  any  section  and  material  may 
be  computed  as  in  Art.  123. 

LOAD  SUSTAINED  BY  THE  SUPPORT. 

For  an  uniformly  distributed  load,  Z,  the  amount  of  vertical 
pressure  that  will  be  sustained  by  each  wall  will  be  obtained 
from  the  formulas  before  deduced  for  a  concentrated  load,  Z, 
at  the  middle  of  the  span,  the  proportion  of  the  total  load 
sustained  by  the  walls  being  the  same  in  each  case  of  loading. 

The  method  usually  given  by  writers  for  determining  the 
load  supported  by  each  wall,  and  the  bending  moment  of  the 
load  applied  to  an  inclined  beam,  is  to  consider  one  half  of  the 


164  STRENGTH   OF   BEAMS   AND    COLUMNS. 

total  load  to  be  sustained  by  each  wall  and  the  bending  mo- 
ment to  be  the  same  as  if  the  beam  were  horizontal,  the  span 
used  in  the  computation  being  the  horizontal  distance,  $, 
between  the  supports.  This  method  of  analysis  is  manifestly 
incorrect,  as  it  would  require  that  an  inclined  beam  of  a  given 
scantling  should  have  a  transverse  strength  that  would  neither 
be  varied  by  its  length  nor  the  angle  that  it  makes  with  the 
horizon,  provided  its  horizontal  span  remained  the  same.  Or 
that  an  inclined  beam  that  is  infinitely  long,  and  a  horizontal 
beam  whose  span  is  the  distance  between  the  supports  of  the 
inclined  beam,  will  have  the  same  transverse  strength,  pro- 
vided the  cross-section  and  material  are  the  same  in  each  beam. 


RECTANGULAR  BEAMS. 


(165) 


8  3  8rfc 

Adding  Eqs.  165  and  166,  we  have,  by  making  C'  +  C"  =  C, 


TRUSSED  BEAMS. 

125.  General  Conditions.  In  trussed  beams  and 
roof -trusses,  beams  are  frequently  subjected  to  both  trans- 
verse and  longitudinal  strains  either  of  compression  or  tension, 
thus  acting  in  the  double  capacity  of  a  horizontal  beam  and  a 
vertical  column. 

RECTANGULAR  BEAMS. 

Let  Fig.  48  represent  such  a  rectangular  beam  loaded 
transversely  with  a  total  load,  Z,  and  longitudinally  with 
either  a  compressive  or  tensile  load,.  Zc. 


COMBINED   BEAMS   AND   COLUMNS. 


165 


From  Eq.  27,  page  38,  in  which  C  —  C",  we  have 


3s 


(168) 
(169) 


And  from  Eq.  99,  page  131,  we  have 

Zc  =  ld,C".  (170) 

Adding  Eqs.  169  and  170,  we. have,  by  making  C'-\-C"—C, 

Sg-*A==**a  (ni) 

mdz 

From  which  either  Z,  Zc,  or  &  may  be  computed  when  the 
other  two  are  known. 

'H-tf-X^  (m) 


3 


(173) 
(174) 


HOLLOW  CIRCULAR  BEAMS. 

The  relation  between  the  transverse  load,  Z,  and  the  result- 
ing moment  of  resistance  of  the  beam  is  given  by  Eq.  91, 
page  62,  in  which  C  —  Cf. 


166  STRENGTH    OF   BEAMS   AND    COLUMNS. 

fc  =  the  factor  from  the  Table,  pages  81  and  104, 
/0  =  the     "          "       "        "        page  136. 


L  =  wftf.0'  =  ,  (175) 


The  compression  from  the  longitudinal  strain,  Zc,  must  be 
equal  to  the  resistance  of  the  beam  as  a  column  from  Eq.  101, 
page  135,  in  which  O  =  C". 

Zc  =  totf.C".  (177) 

Adding  Eqs.  176  and  177  we  have 

Zc  =  %rtf&  (178) 


- 
mrfc 

c«c 

~2/r 

The  compressing  load,  Zc,  in  the  foregoing  formulas,  in 
practice,  generally  results  from  the  transverse  load,  Z,  its 
effect  being  transmitted  to  the  ends  of  the  beam  by  a  vertical 
post  placed  under  the  centre  of  the  beam  and  connected  with 
the  ends  by  inclined  truss-rods.  For  a  single  load,  Z,  at  the 
middle  it  is  usually  the  practice  to  consider  one  half  of  it  as 
producing  a  tensile  strain  on  each  of  the  inclined  truss-rods  ; 
this  is  only  strictly  correct  when  the  beam  is  cut  into  two 
pieces  at  its  centre.  When  the  beam  is  continuous  it  can  only 
strain  the  vertical  post  after  it  begins  to  deflect  ;  therefore  the 
load  that  would  have  caused  the  untrussed  beam  to  deflect 
from  the  horizontal  position  could  bring  no  strain  upon  the 
post. 


INDEX. 


PAGE 

Beams.     See  Transverse  Strength,  Moment  of  Resistance  and  Neutral 
Line. 

Beams,  Cast-Iron,           .        . 64 

"      Wrought- Iron, 87 

"      Steel, 87 

"      Timber, 106 

"      Horizontal,        ..........  3 

"      Inclined . 148 

"      Trussed, 164 

Elastic  Limit 34 

and  Columns,  Combined,          .        .        .        .     '    .        .        .  145 

Bending  Moments, 3 

"        General  Formula,    .         .                 ....  6 

"      Moment  and  Moment  of  Resistance, 22 

Coefficients  of  Strength, 33 

Columns,  Box  Section, 163 

Classed, .        .        .123 

Deflection  of,      .        . 121 

"        Experiments, 122 

"        End  Connections, 124 

Failure  of ,  without  Deflection,       .        .                 .        .        .  125 

"   with              "            127 

"   without  Cantileverage, 130 

"   with                  "                  ...        .        .  137 

General  Conditions  of  Failure, 114 

Gordon's  Formula, 138 

Pin  End, 118 

Resistance  of, 119 

Rectangular  Sections, 128,  131,  137 

Circular  Sections, 129,  133,  140 

Hollow  Circular  Sections,       .        .        .        .        .      130,  135,  141 

"        Angle-Iron,  Eye-Beam,  Channel,  etc.,       ....  143 

Combined  Beams  and  Columns, 145 


168  INDEX. 

PAGE 

Crushing  Strength,  To  Compute 39,  58,  126 

of  Cast-Iron, 64 

To  Compute,      .....  67 

"  Wrought- Iron 87 

To  Compute,       ....  88 

Designing  a  Rectangular  Beam, 39 

"        a  Hodgkinson      "........  44 

a  Double  T           "     .     '    .         .         .         .     '    .         .         .  48 

a  Box                   " 48 

an  Inverted  T      " 53 

a  Double  T          " .53 

Elastic  Limits  Defined,       . 33 

Limit  of  Cast-Iron, .         .         .  65 

"      "  Wrought-Iron, 87 

"      "  Steel,    ..........  88 

"      "  Materials, 33 

Equilibrium,           ...........  2 

Moment  of  Resistance  and  Bending  Moment,     .         .  23 

Factors  of  Safety*          .        .        . 34 

Forces  Denned '.  1 

"     Uniform  Intensity,       .         .         .         .         .         .         .         .         .  2 

"     Uniformly  Varying  Intensity,          ......  2 

Gordon's  Formula, 138 

Inclined  Beams  Supported  at  One  End,      ......  148 

"  Both  Ends,      .        .         .         .    *    .         .160 

Load, 2 

Moment  of  Resistance  Defined 21 

Box  Sections,  General  Formula,    ...  47 

"          Circular  Sections,"  "  .         .         .55 

Double  T  Section,"                          ...  47 

"         "                      Hodgkinson  "         "              "             ...  41 

Hollow  Circular  Sections,  General  Formula,  60 

"        "          "          Eye-Beam  Section,  General  Formula,       .        .  47 

Inverted  T.,                      "  52 

"         "                      Rectangular  Sections,   "              "...  37 

"         "  Concentrated  Parallel  Forces,          .....  3 

"         "  Uniformly  Varying  Forces,         .         .         .         .         .         .  ,8 

Neutral  Line  Defined,          .........  21 

"     Box  Sections,  General  Formula,       .         .         .         .         .48 

"     Cast-Iron  Sections, 78 

"     Wrought-Iron  and  Steel  Sections,     ....  96 

"     Circular  Sections,           .......  56 

"  "       Cast-Iron  Sections,     ....  .80 


INDEX. 

PAGE 

Neutral  Line  Circular  Cast-Iron  Sections,  Table  of,           ...  80 

"       Wrought-Iron  and  Steel  Sections,      .         .         .  100 

"       "       "            "        Table  of,   .  100 

Wooden  Sections, 112 

Table  of,     ....  112 

"           "     Double  T  Section,  General  Formula,       ....  48 

"                       "       "  Wrought-Iron  and  Steel  Sections,         .         .  96 

"  Cast-Iron  Sections, 78 

"     Eye-Beam  Sections, 96 

"           "     Hodgkinson  Section,  General  Formula,    ....  42 

Cast-iron  Section, 75 

"           "                           Wrought-Iron  Section,        ....  96 

"     Hollow  Circular  Sections,     ......  61 

Cast-iron  Sections,        ....  85 

Table  of,      .        .  85 

Wrought-Iron  Sections,         .         .         .  104 

Table  of,      .  104 

"     Inverted  T,  General  Formula,          .....  53 

"          "     Movement  of,         ........  26 

"           "                        "'  from  Experiments,      .....  82 

"           "     Rectangular  Section,  General  Formula,       ...  38 

"                         "            Cast-Iron  Sections,      .....  71 

Table  of,   ...  71 

"                           Wrought-Iron  and  Steel  Sections,       .        .  92 

"       "          "       Table  of,  92 

Wooden  Sections,  Table  of,        ...  108 

"     Position  of,    .........  24 

"     of  Rupture  in  a  Rectangular  Section,      ....  28 

"          "     Transverse  Elastic  Limit,      .        .         .         .         .         .  24 

Notation  for  Rectangular  Sections,       .......  37 

"  Hodgkinson  Beam, 41 

' '  Double  T  and  Box  Beams, 47 

"  Inverted  T  and  Double  Inverted  T,      .  52 

"  Circular  Beams, .55 

"  Hollow  Circular  Beams,        ......  60 

"  Columns 121 

I   "           "  Inclined  Beams,     ........  146 

"  Uniformly  Varying  Forces,      ......  8 

Pin  End  Columns, 118 

Relative  Transverse  Strength  of  Beams, 35 

"         Strength  of  Square  and  Circular  Beams,      ....  59 

"        "          "        Cast-Iron  Beams,        .        .  83 

Timber          "  113 


170  I1STDEX. 

PAGE 

Relative  Value  of  the  Crushing  and  Tensile  Strength,      ...  28 

Resultant  Defined, .2 

"         Uniformly  Varying  Forces,        .        .         .         .         .        8,  16,  19 

Strain, 1 

Stress,         1 

Tensile  Strength,  To  Compute, 39,  58 

"             "         of  Cast-Iron, 64 

To  Compute,                ....  6? 

"    Wrought-Iron  and  Steel,  .        ....  87 

To  Compute 90 

"    Timber, 106 

To  Compute, 106 

Transverse  Strength,  General  Conditions, 21 

"         Box  Beams,  General  Formulas,  ....  48 

"     Cast  Iron  Beams, 78 

"                 "            "     Wrought-Iron  and  Steel  Beams,          .        .  96 

"                  "          Cast-Iron  Beams,     ......  65 

Circular  Beams,  General  Formula,     .        .        .56 

Cast-Iron  Beams,       ....  81 

Wrought-Iron  and  Steel  Beams,  .        .  101 

"                 "                "       Wooden  Beams,        ....  112 

"                 "         Double  T  Beam,  General  Formula,     ...  48 

"                  "               "       "  Wrought-Iron  and  Steel  Beams,      .  96 

"  Cast- Iron  Beams,          .        .         .         .78 

Eye-Beams,  Wrought  Iron  and  Steel,      .        .  96 

"                 "         Hodgkinson  Beam,  General  Formula,        .        .  44 

Cast-Iron  Beam 75 

"    "                 "         Wrought-Iron  and  Steel  Beams,     .  96 

Hollow  Circular  Beams,  General  Formula,     .  61 

"                 "                "            "        Cast-Iron  Beams,      ...  86 

Wrought  Iron  and  Steel  Beams,  104 

Inverted  T,  General  Formula,   .        .                 .  5$ 

Rectangular  Beams,  General  Formula,   .  '      .  38 

"         Cast-Iron  Beams,    ....  71 

"                                       "         Wrought-Iron  and  Steel  Beams,  92 

Wooden  Beams,     .        .        .        .  108 

Steel  Beams 92 

AVrought-Iron  Beams, 91 

Timber  Beams, 106 

Trussed  Beams, '  164 

Working  Load, 34 


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